Module 7 Work, Energy, Power

2005 Pearson Education South Asia Pte Ltd

MECHANICSFE1001 Physics I NTU - College of Engineering

1. Units, Physical Quantities and Vectors

2. Motion Along A Straight Line3. Motion in 2 or 3 Dimensions4. Newton’s Law of Motion

5. Applying Newton’s Laws6. Work and Kinetic Energy7. Potential Energy and Energy Conservation


MECHANICSFE1001 Physics I NTU - College of Engineering

8. Momentum, Impulse, and Collisions

9. Rotation of Rigid Bodies10. Dynamics of Rotational

Motion11. Equilibrium and Elasticity12. Gravitation13. Periodic Motion14. Fluid Mechanics

7. Potential Energy and Energy Conversation


Chapter Objectives

• Understand the importance of potential energy and gravitational potential energy.

• The fundamental principle in all science – the law of conservation of energy.

• Conservation of elastic potential energy that allows us to find the internal energy of the spring.

• See how work-energy theorem explains the transformation from potential to kinetic energy.

• Derive an expression between force and potential energy.• Learn to draw energy diagrams.



Chapter Outline

1. Gravitational Potential Energy2. Elastic Potential Energy3. Conservation and Nonconservative Forces4. Force and Potential Energy5. Energy Diagrams



7.1 Gravitational Potential Energy

• Potential energy is the energy that associated with the position of a system rather than its motion.

• Gravitational potential energy is the potential energy associated with the body’s weight and its height above the ground.

• When a body falls without resistance, work is done by the gravitational force during the vertical motion of a body from an initial height to a final height .

wur

1y2y




• The gravitational potential energy decreases if the body moves downward.




• The gravitational potential energy increases if the body moves upward.



7.1 Gravitational Potential Energy• When a body’s mass m that moves along the (vertical) y-direction, with magnitude and other

forces acting on it; we called the vector sum (resultant) of all the other forces . • With reference from the previous figures, the weight and displacement are in the same direction, so the

work done on the body by its weight is positive; w mg=

otherF

gravW

( )1 2 1 2 (7.1)gravW Fs w y y mgy mgy= = − = −



7.1 Gravitational Potential Energy• If the quantity is positive, will be positive as the weight and displacement is the same

direction and vice versa.• The product of the weight mg and the height y above the origin coordinate is called the gravitational

potential energy, U:

( )1 2y y− gravW

(7.2)U mgy=

(gravitational potential energy)




• We can also express the work done by the gravitational force during the displacement from to as gravW

1y2y

1 2 1 2gravW mgy mgy U U= − = −

( )2 1 (7.3)U U U= − − = −Δ

• The negative sign in front of is essential. • When y increases, work done by gravitational force is negative and the

gravitational potential energy increases, .

U−Δ

0UΔ >



• When y decreases, work done by gravitational force is positive and the gravitational potential energy increases, .

• The unit of potential energy is the joule (J).


0UΔ <

• It is wrong to call the “gravitational potential energy of the body.”• The reason is that gravitational energy is a shared property of the body and the earth.• Note that involve both the body (its mass m) and the earth (the value of g).

U mgy=

U mgy=




Conservation of Mechanical Energy (Gravitational Forces Only)• When a body is falling without air resistance, only

force is the body’s weight acting on it, so .

• Work-energy theorem states that the total work done on the body equals the change in the body’s kinetic energy:

• If gravity is the only force that acts, from Eq. (7.3)

0otherF =ur

2 1totW K K K=Δ = −

1 2tot gravW W U U U= =−Δ = −




Conservation of Mechanical Energy (Gravitational Forces Only)• Putting together we have

K UΔ =−Δ

2 1 1 2K K U U− = −

1 1 2 2 (7.4)K U K U+ = +

2 21 1 2 2

1 1 (7.5)2 2

mv mgy mv mgy+ = +

• If only gravity does work,

or




Conservation of Mechanical Energy (Gravitational Forces Only)• Let the sum of K (kinetic) and U (potential) be E, the

total mechanical energy of the system.

• Eq. (7.4) says that when the body’s weight is the only force doing work on it, .

• The total mechanical energy has the same value at all points during the motion;

1 2E E=

( )E K U= +

constantE K U= + =(if only gravity does work)




Conservation of Mechanical Energy (Gravitational Forces Only)• A quantity that always has the same value is called a

conserved quantity.

• When only the force of gravity does work, the total mechanical energy is constant, that is conserved.

• For example when an athlete jumps, only the gravity does work on him, ignore air resistance.



• Gravitational potential energy does not concern what height we choose, as stays the same.

• Though and depend on where we place the origin, the difference does not change.


Conservation of Mechanical Energy (Gravitational Forces Only)• The total mechanical energy is the same at every

point in the motion, provided that no force other than gravity does work on the body when is it thrown.

2 1y y−

1U 2U

( )2 1 2 1U U mg y y− = −



Example 7.1 Height of a baseball from energy conservation

U throw a 0.145-kg baseball straight up in the air, giving it an initial upward velocity of magnitude 20.0m/s. Use conservation of energy to find how high it goes, ignoring air resistance.



Example 7.1 Height of a baseball from energy conversion



Example 7.1 (SOLN)IdentifyAfter the ball leaves your hand, the only force doing work on the ball is its weight. Hence we can use conservation of mechanical energy.

Set upWe’ll use Eqs.(7.4) and (7.5), taking point 1 to be where the ball leaves your hand and point 2 to be where the ball reached its maximum height. We take the y-axis to point vertically upward. The ball’s speed at point 1 is v1 = 20.0m/s. The ball is instantaneously at rest at the high point of its motion (point 2), so v2 = 0.



Our target variable is how for the ball moves vertically between these two points, so our target variable is the displacement y2 – y1. for simplicity, let’s take the original to be at point 1, where the ball leaves your hand. Then y1 = 0 and the target variable is just y2.

Example 7.1 (SOLN)Set up

sr

ExecuteSince y1 = 0, the potential energy at point 1 is U1 = mgy1 = 0. Furthermore, since the ball is at rest at point 2, the kinetic energy at that point is K2 = ½mv2

2 = 0.



Example 7.1 (SOLN)

sr

ExecuteHence Eq.(7.4), which says that K1 + U1 = K2 + U2, becomes

K1 = U2

As the energy bar graphs shows, the kinetic energy of the ball at point 1 is completely converted into gravitational potential energy at point 2. At point 1 the kinetic energy is

2 21 1

1 1 (0.145 )(20.0 / ) 29.02 2

K mv kg m s J= = =



This equals the gravitational potential energy U2 = mgy2 at point 2, so

We can also solve the equation K1 = U2 algebraically for y2:

Example 7.1 (SOLN)Execute

22 2

29.0 20.4(0.145 )(9.80 / )

U Jy mmg kg m s

= = =

21 2

12

mv mgy=2 21

2 2(20.0 / ) 20.4

2 2(9.80 / )v m sy mg m s

= = =



Example 7.1 (SOLN)EvaluateThe mass divides out, as we should expect; we learned in Chapter 2 that the motion of a body in free fall doesn’t depend on its mass. Indeed, we could have derived the result y2 = v1

2/2g using Eq.(2.13).

In carrying out the calculation above, we chose the origin to be at point 1, so y1 = 0 and U1 = 0. What happens if we make a different choice?



Example 7.1 (SOLN)EvaluateAs an example, suppose we choose the origin to be 5.0m below point 1, so y1 = 5.0m. With this choice the total mechanical energy at point 1 is part kinetic and part potential, while at point 2 it’s purely potential energy. If you work through the calculation again with this choice of origin, you’ll find y2 = 25.4 m; this is 20.4m above point 1, just as with the first choice of origin. In every problem, it’s up to you to choose the height at which U = 0; don’t agonize over the choice, however, because the physics of the answer doesn’t depend on your choice.




Effect of Other Forces

• When additional work is taken into account, the total work done by all other forces is

otherW

(7.6)tot grav otherW W W= +

• Equating kinetic energy,

2 1other gravW W K K+ = −

• From Eq. (7.3), 1 2gravW U U= −

1 2 2 1otherW U U K K+ − = −





• Rearrange we get

1 1 2 2 (7.7)totK U W K U+ + = +

(if forces other than gravity do work)

• Using appropriate expressions,

2 21 1 2 2

1 1 (7.8)2 2totmv mgy W mv mgy+ + = +





• The work done by all forces other than the gravitational force equals the change in the total mechanical energy of the system, where U is the gravitational potential energy.

• When

E K U= +

2 2 1 10, increases, otherW E K U K U> + > +

2 2 1 10, decreases, otherW E K U K U< + < +

0, mechanical energy is constantotherW =




Problem-solving strategy (Problems using mechanical energy)• IDENTIFY

1.Decide whether problems can be solved by energy methods.

2.Energy approach suitable for motion with varying forces, along a curved path, or both.

3.Not suitable for problem involving time.




Problem-solving strategy (Problems using mechanical energy)• SET UP

1.For energy method, decide what the initial and final velocities and positions of the system.

2.Define coordinate system, we suggest position y-direction upward.

3.Draw free-body diagram and identify all non-gravitational forces that do work.

4.List all the unknown and known quantities and decide which is the target variables.




Problem-solving strategy (Problems using mechanical energy)• EXECUTE

1.Write expressions involving for the initial and final kinetic and potential energies.

2.Relate kinetic, potential and gravitational energies to using Eq. (7.7).

3.Useful to draw bar charts showing the initial and final values of

1 2 1 2, , and K K U U

otherW

, and .K U E K U= +




Problem-solving strategy (Problems using mechanical energy)• EVALUATE

1.Check whether your answers makes physical sense.

2.The work done by each force must be represented either or as , but never in both places.

3.The gravitational work included in , make sure you did not include it again in .

1 2U U U− =−Δ otherW

otherW



Example 7.2 Work and energy throwing a baseball

In Example 7.1, suppose your hand moves up 0.50m while you are throwing the ball, which leaves your hand with an upward velocity of 20.0 m/s. Again ignore air resistance. a) Assuming that your hand exert a constant upward force on the ball, find the magnitude of that force. b) Find the speed of the ball at a point 15.0m above the point where it leaves your hand.



Example 7.2 (SOLN)IdentifyIn Example 7.1 we used conservation of mechanical energy because only gravity did work. In this example, however, we must also include the nongravitational work done by your hand.



Example 7.2 (SOLN)



Example 7.2 (SOLN)Set upFigure (a) shows a diagram of the situation, including a FBD for the ball while it is being thrown. The ball’s motion occurs in two stages: while it is in contact with your hand and after it leaves your hand. To keep track of these stages, we let point 1 be where your hand first starts to move, point 2 be where the ball leaves your hand, and point 3 be where the ball is 15.0m above point 2. The nongravitational force of your hand acts only between points 1 and point 2.

Fr



Example 7.2 (SOLN)Set upUsing the same coordinate system as in Example 7.1, we have y1 = - 0.50 m, y2 = 0 and y3 = 15.0 m. The ball starts at rest at point 1, so v1 = 0, and we are given that the ball’s speed as it leaves your hand is v2 = 20.0m/s. Our target variables are (a) magnitude F of the force of your hand and (b) the speed v3 at point 3.



a)To determine the magnitude of , we’ll first use Eq.(7.7) to calculate the work Wother done by this force, we have

K1 = 0.


Fr

21 1 (0.145 )(9.80 / )( 0.50 ) 0.71U mgy kg m s m J= = − =−

2 22 2

1 1 (0.145 )(20.0 / ) 29.02 2

K mv kg m s J= = =

22 2 (0.145 )(9.80 / )(0) 0U mgy kg m s= = =



Example 7.2 (SOLN)ExecuteThe initial potential energy U1 is negative because the ball initially below the origin. According to Eq.(7.7)

, so

2 1 2 1( ) ( )otherW K K U U= − + −

(29.0 0) (0 ( 0.71 )) 29.7J J J= − + − − =

The kinetic energy of the ball increase by K2 – K1 = 29.0J, and the potential energy increase by U2 – U1 = 0.71J; the sum is E2 – E1, the change in total mechanical energy, which is equal to Wother.

1 1 2 2otherK U W K U+ + = +



This is about 40 times greater than the weight of the ball.

Example 7.2 (SOLN)ExecuteAssuming that the upward force that your hand applies is constant, the work Wother done by this force is equal to the magnitude F of the force multiplied by the upward displacement y2 – y1 over which it acts:

2 1( )otherW F y y= −

2 1

29.7 590.50

otherW JF Ny y m

= = =−

Fr



Example 7.2 (SOLN)Executeb) To find the speed at point 3, note that between point 2 and point 3, total mechanical energy is conserved; the force of your hand no longer acts, so Wother = 0. we can then find the kinetic energy at point 3 using Eq.(7.4):

2 2 3 3K U K U+ = +2

3 3 (0.145 )(9.80 / )(15.0 ) 21.3U mgy kg m s m J= = =

3 2 2 3( ) (29.0 0 ) 21.3 7.7K K U U J J J J= + − = + − =



Example 7.2 (SOLN)ExecuteSince , where is the y-component of the ball’s velocity at point 3, we have

23 3

12 yK mv= 3yv

33

2 2(7.7 ) 10 /0.145y

K Jv m sm kg

=± =± =±

The significant of the plus-or-minus sign is that the ball passes point 3 twice, once on the way up and again on the way down. The total mechanical energy E is constant and equal to 29.0J while the ball is in free fall, and the potential energy at point 3 is U3 = 21.3J whether the ball is moving up or down.




So at point 3, the ball’s kinetic energy K3 and speed don’t depend on the direction the ball is moving. The velocity v3y is positive (+10m/s) when the ball is moving up and negative (10m/s) when it is moving down; the speed v3 is 10 m/s in either case.



Example 7.2 (SOLN)Evaluate

As a check on our result, recall from Example 7.1 that the ball reaches a maximum height y = 20.4m. At that point all of the kinetic energy that the ball had when it left your hand at y = 0 has been converted to gravitational potential energy. At y = 15.0m, the ball is about three-fourths of its mechanical energy should be in the form of potential energy. Can you show that this is true from our results for K3 and U3?




Gravitational Potential Energy for Motion Along a Curved Path• When a body moves along a path which is slanted as

shown, the body is acted on by the gravitational force and possibly by other forces we called .

w mg=ur ur

otherFur




Gravitational Potential Energy for Motion Along a Curved Path• To find the work done by gravitational force during

this displacement, we divided the path up into small segment . sΔ

r




Gravitational Potential Energy for Motion Along a Curved Path• In terms of unit vectors, the force is

and the displacement is

• The work done is

• Every segment has the same work done by the gravity.

• Thus the total work done is

$w mg mg j= =−ur ur

$.s xi y jΔ =Δ +Δr $

$ $( ) .w s mg j xi y j mg y⋅Δ = − ⋅ Δ + Δ = − Δur r $

( )2 1 2 1 1 2gravW mg y y mgy mgy U U= − − = − = −




Gravitational Potential Energy for Motion Along a Curved Path• This work is unaffected by any horizontal motion that

may occur.

• We can use the same expression for gravitational potential energy whether the body’s path is curved or straight.



Example 7.3 Energy in projectile motion

A batter hits two identical baseballs with the some initial speed and height but different initial angles. Prove that at a given height h, both balls have the same speed of air resistance can be neglected.



Example 7.3 (SOLN)



If there is no air resistance, the only force acting on each ball after it is hit is its height. Hence the total mechanical energy for each ball is constant. Figure shows the trajectories of two balls batted at the same height with the same initial speed, and thus the same total mechanical energy, but with different initial angles. At all points at the same height the potential energy is the same. Thus the kinetic energy at this height must be the same for both balls, and the speed are the same.

Example 7.3 (SOLN)



Example 7.4 Maximum height of a projectile using energy methods

In Example 3.10 (Section 3.3) we derived an expression for the maximum height h of a projectile launched with initial speed vo at initial angle ao:

Derive this expression using energy considerations.

2 2sin2

o ovh

gα

=



Example 7.4 (SOLN)



Example 7.4 (SOLN)

We neglect air resistance, so just as in Example 7.3, total mechanical energy is conserved. Let point 1 at y = 0 be the launch point, where the speed is v1 = v0, and let point 2 at y = h be the highest point on the trajectory. Our target variable is the maximum height h, at which the kinetic energy is minimum and the gravitational potential energy is maximum. This appears to be an easy problem: the potential energy at point 2 is U2 = mgh, so it may seem that all we need to do is solve the energy-conservation equation K1 + U1 = K2 + U2 for U2.

Identify and Set up



Example 7.4 (SOLN)

However, while we know the initial kinetic and potential energies (K1 = ½ mv1

2 = ½ mv02 and U1 = 0),

we don’t know the speed or kinetic energy at point 2. to get around this issue, we’ll use two results from our study of projectile motion in Chapter 3: (1) the x-component of acceleration is zero, so the x-component of velocity is constant, and (2) the y-component of velocity is zero at point 2 (the highest point of the trajectory).

Identify and Set up



Example 7.4 (SOLN)ExecuteWe can express the kinetic energy at each point in terms of the components of velocity, using v2 = vx

2 + vy

3: 2 21 1 1

1 ( )2 x yK m v v= +

2 22 2 2

1 ( )2 x yK m v v= +

Conservation of energy then gives K1 + U 1 = K2 + U2, so

2 2 2 21 1 2 2

1 1( ) 0 ( )2 2x y x ym v v m v v mgh+ + = + +



Example 7.4 (SOLN)ExecuteTo simplify this, we multiply through by 2/m to obtain

Now we use results from projectile motion. Since the x-component of velocity doesn’t change, v1x = v2x and we can cancel the vx

2 terms from both sides of the equation given above. Furthermore, because the projectile has zero vertical velocity at the highest point of its motion, v2y = 0. Hence we get

2 2 2 21 1 2 2 2x y x yv v v v gh+ = + +

21 2yv gh=



Example 7.4 (SOLN)ExecuteBut v1y is just the y-component of initial velocity, which is equal to . Making this substitution and solving for h, we find

2 20 0sin

2v

hg

α=

This agrees with the result of Example 3.10, as it must.

0 0sinv α

Evaluate



Your cousin Throckmorton skateboards down a curved playground ramp. Treating Throcky and his skateboard as a particle, he moves through a quarter-circle with radius R. The total mass of Throcky and his skateboard is 25.0kg. He starts from rest and there is no friction. (a) Find his speed at the bottom of the curve.

Example 7.5 Calculating speed along a vertical circle



Example 7.5



We can’t use the equations of motion with constant acceleration isn’t constant because the slope decreases as Throcky descend. Instead, we’ll use the energy approach. Since Throcky moves along a circular arc, we’ll also use what we learned about circular motion is Section 5.4.

Example 7.5 (SOLN)Identify



Since there is no friction, the only force other than Throcky’s weight is the normal force exerted by the ramp (Fig. b). Although this force acts all along the path, it does zero work because is perpendicular to Throcky’s velocity at every point. Hence Wother = 0 and mechanical energy is conserved.


nr

nr



Take point 1 at the starting point and point 2 at the bottom of the curved ramp, and let y = 0 be at the bottom of the ramp. Then y1 = R and y2 = 0. (We are treating Throcky as if his entire mass were concentrated at his center.) Throcky starts at rest at the top, so v1 = 0. Our target variable in part (a) is his speed at the bottom, v2. In part (b) we want to find the magnitude n of the normal force at point 2. As this force does no work, it doesn’t appear in the energy equation, so we’ll use Newton’s second law instead.




Example 7.5 (SOLN)Executea)The various energy quantities are

K1 = 0 U1 = mgR

K2 = ½ mv22 U2 = 0

From conservation of energy,

K1 + U1 = K2 + U2

2 2v gR=

22

10 02

mgR mv+ = +



The speed is the same as if Throcky had fallen vertically through a height R, and it is independent of his mass. As a numerical example, let R = 3.00 m. Then

Notice that this answer doesn’t depend on the ramp being circular; no matter what the shape of the ramp. Throcky will have the same speed at the bottom. This would be true even if the wheels of his skateboard lost contact with the ramp during the ride, because only the gravitational force would still do work.


22 2(9.80 / )(3.00 ) 7.67 /v m s m m s= =

2 2v gR=



b) To find n at point 2 using Newton’s second law, we need the FBD at that point. Fig.b. At point 2, Throcky is moving at speed in a circle of radius R; his acceleration is toward the center of the circle and has magnitude

If we take the positive y-direction to be upward, the u-component of Newton’s second law is


22 2 2rad

v gRa gR R

= = =

2 2v gR=

( ) 2radyF n w ma mg= + − = =∑2 3n w mg mg= + =



At point 2 the normal force is three time Throcky’s weight. This result is independent of the radius of the circular ramp. We learned in Example 5.10 (Section 5.2) and Example 5.25 (Section 5.4) that the magnitude of n is the apparent weight, so Throcky feels as though he weighs three times his true weight mg. But as soon as he reaches the horizontal part of the ramp to the right of point 2, the normal force decreases to w = mg and Throcky feels normal again. Can you see why?




This example shows a general rule about the role of forces in problems in which we use energy techniques: What matters is not simply whether a force acts, but whether that force does work. If does not appear at all in Eq.(7.7), K1 + U1 +Wother = K2 + U2.

Notice we had to use both the energy approach and Newton’s second law to solve this problem; energy conservation gave us the speed and gave us the normal force. For each part of the problem we used the technique that most easily led us to the answer.


F ma=∑r r



In Example 7.5, suppose that the ramp is not frictionless and that Throcky’s speed at the bottom is only 6.00 m/s. What work was done by the friction force acting on him? Use R = 3.00m.

Example 7.6 A vertical circle with friction

We use the same coordinate system and the same initial and the same initial and final points as in Example 7.5. Again the normal force does no work, but now there is a friction force that does do work. Hence the nongravitational work done on Throcky between points 1 and 2, Wother, is just equal to the work done by friction, Wf. This is our target variable, which we’ll find using Eq.(7.7).

Identify and Set up

fr



Example 7.6 (SOLN)



Example 7.6 (SOLN)

The energy quantities are

K1 = 0

U2 = 0

From Eq.(7.7),

Execute

2 22 2

1 1 (25.0 )(6.00 / ) 4502 2

K mv kg m s J= = =

21 (25.0 )(9.80 / )(3.00 ) 735U mgR kg m s m J= = =

2 2 1 1fW K U K U= + − −

450 0 0 735 285J J J= + − − =−



Example 7.6 (SOLN)

The work done by the friction force is -285J, and the total mechanical energy decreases by 285J. Do you see why Wf has to be negative?

Execute

EvaluateThrocky’s motion is determined by Newton’s second law . But it would be very difficult to apply the second law directly to this problem because the normal and friction forces and the acceleration are continuously changing in both magnitude and direction as Throcky skates down. The energy approach, by contrast

F ma=∑r r



Example 7.6 (SOLN)EvaluateThe energy approach, by contrast, relates the motion at the top and bottom of the ramp without involving the details of what happens in between. Many problems are easy if energy considerations are used but very complex it we try to use Newton’s laws directly.



Example 7.7 An inclined plane with frictionWe want to load a 12-kg crate into a truck by sliding it up a ramp 2.5m long, inclined at 30o. A worker, giving no thought to friction, calculates that he can get the crate up the ramp by giving it an initial speed of 5.0m/s at the bottom and letting it go. But friction is not negligible; the crate slides 1.6m up the ramp, stops, and slides back down. A) Assuming that the friction force acting on the crate is constant, find its magnitude. b) How fast is the crate moving when it reaches the bottom of the ramp?



Example 7.7 An inclined plane with friction



Example 7.7 (SOLN)

The friction force does work on the crate as it slides. As in Example 7.2, we’ll use the energy approach in part (a) to find the magnitude of the nongravitational force that does work (in this case, friction). Once we know the magnitude of the friction force, we can calculate how much nongravitational work this force does as the crate slides back down. We can then use the energy approach again to find the crate’s speed at the bottom of the ramp.

Identify



Example 7.7 (SOLN)

The first part of the motion is from point 1, at the bottom of the ramp, to point 2, where the crate stops instantaneously. In the second part of the motion, the crate returns to the bottom of the ramp, which we’ll call point 3. We’ll take y = 0 (and hence U = 0) to be at ground level, so y1 = 0, y2 = (1.6m)sin30o = 0.80m, and y3 = 0. We are given that v1 = 5.0m/s and v2 = 0 (the crate is instantaneously at rest at point 2). Our target variable in part (a) is f, the magnitude of the friction force, which we’ll find using Eq.(7.7). In part (b) our target variable is v3, the speed at the bottom of the ramp.

Set up



Example 7.7 (SOLN)

a) The energy quantities areExecute

21

1 (12 )(5.0 / ) 1502

K kg m s J= = 1 0U =

2 0K = 22 (12 )(9.8 / )(0.80 ) 94U kg m s m J= =

otherW fs=−

Here is the unknown magnitude of the friction force and s = 1.6m. Using Eq.(7.7), we findf

1 1 2 2otherK U W K U+ + = +

2 2 1 1( ) ( )otherW fs K U K U=− = + − +



Example 7.7 (SOLN)

The friction force of 35N, acting over 1.6m, causes the mechanical energy of the crate to decrease from 150J to 94J (Fig.c).

Execute2 2 1 1( ) ( )K U K U

fs

+ − +=

[ ](0 94 ) (150 0)35

1.6J J

Nm

+ − += =



Example 7.7 (SOLN)

b) The crate returns to point 3 at the bottom of the ramp; y3 = 0 and U3 = 0 (Fig.b). On the way down, the friction force and the displacement both reverse direction but have the same magnitudes, so the friction work has the same negative value for each half of the trip. The total work done by friction between points 1 and 3 is

Execute

2 2(35 )(1.6 ) 112other fricW W fs N m J= =− =− =−



Example 7.7 (SOLN)

From part (a), K1 = 150J and U1 = 0. Equation (7.7) then gives

Execute

1 1 3 3otherK U W K U+ + = +

3 1 1 3 otherK K U U W= + − +150 0 0 ( 112 ) 38J J J= + − + − =

The crate returns to the bottom of the ramp with only 38J of the original 150J of mechanical energy (Fig.c). Using K3 = ½ mv3

2, we get

32(38 ) 2.5 /12

Jv m skg

= =



Example 7.7 (SOLN)

Notice that the crate’s speed when it returns to the bottom of the ramp, v3 = 2.5 m/s, is less than the speed v1 = 5.0 m/s at which it left that point. That’s good – energy was lost due to friction.

Notice also that in part (b) we applied Eq.(7.7) to point 1 and 3, considering the entire round trip as a whole. Alternatively, we could have considered the second part of the motion by itself and applied Eq.(7.7) to point 2 and 3. Try it and see whether you get the same result for v3.

Evaluate



7.2 Elastic Potential Energy• Elastic potential energy is the storing energy in a body.• A body is said to be elastic if it returns to its original shape and size after being deformed, e,g. a spring.• To stretch a spring by a distance x, we need to exert a force , where k is the force constant of the

spring and provided that x is sufficiently small.

F kx=



7.2 Elastic Potential Energy• The figure shows an ideal spring attached to a mass m moving along x-direction.• The body is at when the spring is neither stretched nor compressed.

0x =



7.2 Elastic Potential Energy

• When the block is stretched to a distance and released, how much work is done from to ? 1x 2x




• Therefore from Section 6.3, we find that the work done on the spring is

2 22 1

1 12 2

W kx kx= −

where k is the force constant of the spring.• Stretching of spring is doing positive work on the spring and releasing is do

negative work.• If we need to find the work done by the spring,

2 21 2

1 12 2elW kx kx= −

where “el” stands for elastic.




• When and are positive and , the spring does negative work.

• When and are positive and , the spring does positive work.

• We define elastic potential energy as:

1x 2x 2 1x x>

1x 2x 2 1x x<

21 (7.9)2

U kx=

(elastic potential energy)

• The unit of U is joule (J), same unit for all energy and work quantities.




• Using Eq. (7.9) to express the work done by the elastic force in terms of the change in potential energy:

2 21 2 1 2

1 1 (7.10)2 2elW kx kx U U U= − = − =−Δ

• When spring is stretched, is negative and U increases. • When spring relaxes, is negative and U decreases, the

spring loses elastic potential energy.

elW

elW



7.2 Elastic Potential Energy• The graph of elastic potential energy for an ideal spring is a parabola:

, where x is the extension or compression of the spring.• For extension (stretching), x is positive.• For compression (when that is possible), x is negative. • Elastic potential energy U is never negative.




• Difference between and is that we do not have the freedom to choose to be wherever we wish.

• To be consistence, must be the position where the spring is neither stretched nor compressed.

U mgy= 212

U kx=

0x =0x =




• If elastic force is the only force that does work on the body, then

1 2tot elW W U U= − −

• The work-energy theorem then gives us

1 1 2 2 (7.12)K U K U+ = +

2 1totW K K= −

(if only the elastic force does work)




• U is given by Eq. (7.9), so

2 2 2 21 1 2 2

1 1 1 1 (7.12)2 2 2 2

mv kx mv kx+ = +

(if only the elastic force does work)

• In this case, the total mechanical energy (the sum of kinetic and elastic potential energy) is conserved.

E K U= +




• All along we are only discussing ideal spring that is massless in order Eq. (7.12) to be correct.

• We neglect the mass of spring if its mass is less than the mass m of the body attached to the spring.

• If forces other than elastic force also do work on the body, then the total work is

tot el otherW W W= +

2 1el otherW W K K+ = −




• The work done by the spring is still .• So again,

1 2elW U U= −

1 1 2 2 (7.13)otherK U W K U+ + = +

(if forces other than the elastic force does work)

and

2 2 2 21 1 2 2

1 1 1 1 (7.14)2 2 2 2othermv kx W mv kx+ + = +

(if forces other than the elastic force does work)




• The equation shows that the work done by all forces other than the elastic force equals the change in the total mechanical energy of the system, where U is the elastic potential energy.

• When is positive, E increases. When is negative, E decreases.E K U= +

otherW otherW




Situations with Both Gravitational and Elastic Potential Energy• When we have both gravitational and elastic energy,

we have .

• In general form,

• The work done by all forces other than the gravitational force or elastic force equals the change in the total mechanical energy E = K + U of the system, where U is the sum of the gravitational potential energy and the elastic potential energy.

grav elU U U= +

1 ,1 2 ,2 ,2 (7.15)grav el other grav elK U U W K U U+ + + + + +




Situations with Both Gravitational and Elastic Potential Energy• If only gravitational and elastic forces are acting on

the body, the total mechanical energy is conserved.

• The work done by the gravitational and elastic force is accounted for by their potential energies; the work of the other forces, has to be included separately.

otherW



Example 7.8 Motion with elastic potential energyIn Fig.a shown in the nest slide, a glider with mass m = 0.200kg sits on a frictionless horizontal air track, connected to a spring with force constant k = 5.00N/m. You pull on the glider, stretching the spring 0.100m, and then release it with no initial velocity (Fig. b). The glider begins to move back toward its equilibrium position (x = 0). What is its x-velocity when x = 0.080m?



Example 7.8 Motion with elastic potential energy



Example 7.8 (SOLN)

Because the spring force varies with position, this problem can’t be solved with the equations for motion with constant acceleration. Instead, we’ll use the energy method to find a simple solution. In particular, we’ll use the idea that as the glider starts to move, elastic potential energy is converted into kinetic energy. (The glider remains at the same height throughout the motion, so gravitational potential energy is not a factor.)

Identify



Example 7.8 (SOLN)

The spring force is the only force doing the work on the glider, so Wother = 0 and we may use Eq.(7.11). Let point 1 be where the glider is released (Fig. b), and let point 2 be at x = 0.080m variable is the x-velocity at point 2, v2x.

Set up



Example 7.8 (SOLN)

The energy quantities areExecute

21

1 (0.200 )(0) 02

K kg= =

21

1 (5.00 / )(0.100 ) 0.02502

U n m m J= =

22 2

12 xK mv=

22

1 (5.00 / )(0.080 ) 0.01602

U N m m J= =



Example 7.8 (SOLN)ExecuteThen from Eq. (7.11),

2 1 1 2 0 0.0250 0.0160 0.0090K K U U J J J= + − = + − =

22

2 2(0.0090 ) 0.30 /0.200x

K Jv m sm kg

=± =± =±

We choose the negative root because the glider is moving in the – x-direction; the answer we want is v2x = -0.30m/s.



Example 7.8 (SOLN)EvaluateWhat is the meaning of the second solution, v2x = +0.30m/s? Eventually the spring will compress and push the glider back to the right in the positive x-direction. (see Fig.d). The second solution tells us that when the glider passes through x = 0.080m while moving to the right, its speed will be 0.30m/s – the same speed as when it passed through this point while moving to the left. When the glider passes through the point x = 0, the spring is relaxed and all of the mechanical energy is in the form of kinetic energy. Can you show that the speed of the glider at this point is 0.50m/s?



Example 7.9 Motion with elastic potential energy and work done by other forces

For the system of Example 7.8, suppose the glider is initially at rest at x = 0, with the spring unstretched. You then apply a constant force in the + x-direction with magnitude 0.610N to the glider. What is the glider’s velocity when it has moved to x = 0.100m?

Fr



Example 7.9 (SOLN)

Although the force you apply is constant, the spring force isn’t, so the acceleration of the glider won’t be constant. Total mechanical energy is not conserved because of the work done by the force , but we can still use the energy relation as given by Eq.(7.13).

IdentifyFr

Fr

Let point 1 be at x = 0, where the velocity is v1x = 0, and let point 2 be at x = 0.100m. (These points are different from the ones labeled in figure in Example 7.8.) our target variable is v2x, the velocity at point 2.

Set up



Example 7.9 (SOLN)

The energy quantities areExecute

1 0K = 1 0U =2

2 212 xK mv= 2

21 (5.00 / )(0.100 )2

U N m m=

0.0250J=(0.610 )(0.100 ) 0.0610otherW N m J= =

(To calculate Wother we multiplied the magnitude of the force by the displacement, since both are in the + x-direction.) Initially, the total mechanical energy is zero; the work done by the force increases the total mechanical energy to 0.0610J, of which 0.0250J is elastic potential energy.

Fr



The remainder is kinetic energy. From Eq.(7.13),


1 1 2 2otherK U W K U+ + = +

2 1 1 2otherK K U W U= + + −

0 0 0.0610 0.0250 0.0360J J J= + + − =

22

2 2(0.0360 ) 0.60 /0.200x

K Jv m sm kg

= = =

We choose the positive square root because the glider is moving in the + x-direction.



Example 7.9 (SOLN)

To test our answer, think what would be different if we disconnected the glider from the spring. Then would be the only force doing work, there would be zero potential energy at all times, and Eq.(7.13) would give us

Evaluate

Fr

2 1 0 0.0610otherK K W J= + = +

22

2 2(0.0610 ) 0.78 /0.200x

K Jv m sm kg

= = =

We found a lower velocity than this value that use the spring does negative work on the glider as it stretches (Fig. 7.13b in Example 7.8.)



Example 7.10 Motion with elastic potential energy after other forces have ceased

In Example 7.9, suppose the force is removed when the glider reaches the point x = 0.100m. How much farther does the glider move before coming to rest?

Fr

After is removed, the spring force is the only force doing work. Hence for this part of the motion the mechanical energy E = K + U is conserved.

IdentifyFr



Example 7.10 (SOLN)

We’ll let point 2 be at x = 0.100m, as in Example 7.9, and let point 3 be where the glider comes instantaneously to rest. Our target variable is the coordinate x3 of this point. We’ll find its value using the conservation of energy expressions, Eq.(7.11), along with the relation U = ½ kx2 for elastic potential energy.

Set up



Example 7.10 (SOLN)

We saw in Example 7.9 that the kinetic and potential energies at point 2 are K2 = 0.0360J and U2 = 0.0250J. The total mechanical energy at and beyond this point is therefore K2 + U2 = 0.0610J. When the glider comes to rest at x = x3, the kinetic energy K3 is zero and the elestic potential energy U3 is equal to the total mechanical energy 0.0610J. We can also see this from K2 + U2 = K3 + U3:

U3 = K2 + U2 – K3 = 0.0360J + 0.0250J – 0 0.0610J

Execute



Example 7.10 (SOLN)

But , so Execute

23 3

12

U kx=

33

2 2(0.0610 ) 0.1565.00 /

U Jx mk N m

= = =

The body moves an additional 0.056 m after the force is removes at x = 0.100m.



Example 7.10 (SOLN)

The total mechanical energy for the motion from point 2 to point 3 is 0.0610J, the same as the work Wother done by the force in Example 7.8. Is this just coincidence? Not at all, the system of glider and spring had zero mechanical energy initially (at point 1 in Example 7.9), so all the mechanical energy it has came from the work done by .

Evaluate

Fr

Fr



Example 7.11 Motion with gravitational, elastic, and friction forces

In a “worst-case” design scenario, a 2000-kg elevator with broken cables is falling at 25 m/s when it first contacts a cushioning spring at the bottom of the shaft. The spring is supposed to stop the elevator, compressing 3.00m as it does so. During the motion a safety clamp applies a constant 17,000-N frictional force to the elevator. As a design consultant, you are asked to determine what the force constant of the spring should be.



Example 7.11



Example 7.11 (SOLN)

We’ll use the energy approach to determine the force constant, which appears in the expression for elastic potential energy. Furthermore, total mechanical energy is not conserved because the friction force does negative work Wother on the elevator.

Identify



Example 7.11 (SOLN)

Since mechanical energy isn’t conserved and more than one kind of potential energy is involved, we’ll use the most general form of the energy relation, Eq.(7.15). We take point 1 as the position of the bottom of the bottom of the elevator when it initially contacts the spring, and take point 2 as its position when it is at rest. We choose the origin to be at point 1, so y1 = 0 and y2 = -3.00m. With this choice the coordinate of the upper end of the spring is the same as the coordinate of the elevator, so the elastic potential energy at any point between point 1 and point 2 is Uel = ½ky2. ( The gravitational potential energy is Ugrav = mgy as usual.)

Set up



Example 7.11 (SOLN)

We know the initial and final speeds of the elevator and the magnitude of the friction force, so the only unknown is the force constant k (our target variable).

Set up

The elevator’s initial speed is v1 = 25m/s, so the initial kinetic energy is

Execute

2 21

1 1 (2000 )(25 / ) 625,0002 2

K mv kg m s J= = =



Example 7.11 (SOLN)ExecuteThe elevator stop at point 2, so K2 = 0. The potential energy at point 1, U1, is zero; Ugrav is zero because y1 = 0, and Uel = 0 because the spring is not yet compressed. At point 2 there is both gravitational and elastic potential energy, so

22 2 2

12

U mgy ky= +

The gravitational potential energy at point 2 is2

2 (2000 )(9.80 / )( 3.00 ) 58,800mgy kg m s m J= − =−



Example 7.11 (SOLN)

The other force is the 17,000-N friction force, acting opposite to the 3.00-m displacement, so

Execute

Putting these term into , we have 2

1 2 210 0 ( )2otherK W mgy ky+ + = + +

(17,000 )(3.00 ) 51,000otherW N m J=− =−

1 2 20 otherK W K U+ + = +



Example 7.11 (SOLN)ExecuteSo the force constant of the spring is

1 222

2( )otherK W mgyk

y+ −

=

[ ]2

2 625,000 ( 51,000 ) ( 58,800 )

( 3.00 )

J J J

m

+ − − −=

−51.41 10 /N m= ×

This is comparable to the spring in an automobile suspension



Example 7.11 (SOLN)

Let’s note what might seem to be a paradox in this problem. The elastic potential energy in the spring at point 2 is

Evaluate

2 5 22

1 1 (1.41 10 / )( 3.00 ) 632,8002 2

ky N m m J= × − =

This is more than the total mechanical energy at point 1,

1 1 1 625,000 0 625,000E K U J J= + = + =



Example 7.11 (SOLN)EvaluateBut the friction force caused the mechanical energy of the system to decrease by 51,000J between point 1 and point 2. Does this mean that energy appeared from nowthere? Don’t panic; there is no paradox. At point 2 there is also negative gravitational potential energy mgy2 = -58,800J because point is below the origin. The total mechanical energy at point 2 is

22 2 2 2 2

102

E K U ky mgy= + = + +

632,800 ( 58,800 ) 574,000J J J= + − =



Example 7.11 (SOLN)EvaluateThis is just the initial mechanical energy of 625,000J, minus 51,000J lost to friction.

Your next jobs as design consultant would be to tell your clients that the elevator won’t stay at the bottom of the shaft. Instead, it will bounce back up. The reason is that at point 2 the compressed spring exerts an upward force of magnitude Fspring = (1.41x 105 N/m)(3.00m) = 422,000N. The weight of the elevator is only w = mg = (200kg)(9.80m/s2) = 19,600N, so the net force will be upward.



Example 7.11 (SOLN)EvaluateThe elevator bounces back upward even if the safety clamp now exerts a downward friction force of magnitude f = 17,000N; the spring force is greater than the sum of f and mg. The elevator will spring bounce again and again until enough mechanical energy has been removed by friction for it to stop.

Can you also show that the acceleration of the falling elevator when it hits the spring is unacceptably high?



7.3 Conservative and Nonconservative Forces

• A force that conserve the amount of energy between kinetic and potential energies is called a conservative force.

• For any conservative force the work done by that force depends on the end points, not on the path taken.

• From the diagram, the gravitational force is conservative and it does the same work on the runner no matter what path he takes from point 1 to point 2, the mechanical energy is constant.







• The work done by a conservative force always has these properties:1. It can always be expressed as the difference between the initial and final values of a potential energy function.2. It is reversible.3. It is dependent of the path of the body and depends only on the starting and ending points.4. When the starting and ending points are the same, the total work is zero.




• When the only forces that do work are conservative forces, the total mechanical energy is constant.

• A force that is not conservative is called nonconservative force. • The work done by a nonconservative force cannot be represented by a potential-energy

function and can also cause mechanical energy to be lost or dissipate, which is called dissipative force.

E K U= +



Example 7.12 Friction work depends on pathYou are rearranging your furniture and wish to move a 40.0kg futon 2.50 m across the room. However, the straight-line path is blocked by a heavy coffee table that you don't want to move. Instead, you slide the futon in a dogleg path over the floor; the doglegs are 2.00 m and 1.50 m long. Compared to the straight-line path, how much work must you do to push the futon in the dogleg path? The coefficient of kinetic friction is 0.200.



Example 7.12 Friction work depends on path



Example 7.12 (SOLN)

Here work is done both by you and by the force of friction, so we must use the energy relation that includes forces other than elastic or gravitational forces. We’ll use this relation to find a connection between the work that you do and the work done by friction.

Identify



Example 7.12 (SOLN)

Figure shows the initial and final points. The futon at the rest at both point 1 and point 2, so K1 = K2 = 0. The gravitational potential energy does not change because the futon moves only horizontally; to be specific, we’ll say U1 = U2 = 0. From Eq.(7.7) it follows that Wother = 0. The other work done on the futon is the sum of the positive work you do, Wyou, and the negative work Wfric done by the kinetic friction force. Since the sum of these is zero, we have Wyou = -Wfric. Thus to determine Wyou, we’ll calculate the work done by friction.

Set up



Example 7.12 (SOLN)

Because the floor is horizontal, the normal force on the futon equals its weight mg, and the magnitude of the friction force is . The work you must do over each path id then

Execute

k k kf n mgμ μ= =

( )you fric k kW W f s mgsμ=− =−− =+2(0.200)(40.0 )(9.80 / )(2.50 )kg m s m=

196J= (straight-line path)



Example 7.12 (SOLN)

The extra work you must do is 274J – 196J = 78J

Executeyou fricW W=−

2(0.200)(40.0 )(9.80 / )(2.00 1.50 )kg m s m m= +

274J= (dogleg path)

EvaluateThe work done by friction is Wfric = -Wyou = -196J on the straight-line path, and -274J on the dogleg. The work done by friction depends on the path taken, which illustrates that friction is a nonconservative force.



In a certain region of space on an election is , where C is a positive constant. The election moves in a counter-clockwise direction around a square loop in the xy-plane. The corners of the square are at (x,y) = (0,0), (L,0), (L,L) and (0,L). Calculate the work done on the electron by the force during one complete trip around the square. Is this force conservative or nonconservative?

Example 7.13 Conservative or nonconservative?ˆxF C j=

r

Fr



Example 7.13 Conservative or nonconservative?



In Example 7.12 the force of the friction was constant in magnitude and always opposite to the displacement, so it was easy to calculate the work done. Here, however, the force is not constant and in general is not in the same direction as the displacement. So we’ll use the more general expression for work, Eq.(6.14):

Example 7.13 (SOLN)Identify and Set up

Fr

2

1

P

pW F dl=∫

rrg

Where is an infinitesimal displacement. dlr



Let’s calculate the work done by the force on each leg of the square and then add the results to find the work done on the round trip.


Fr

On the first leg, from (0,0) to (L,)), the force varies but is everywhere perpendicular to the displacement. So , and the work done on the first leg is W1 = 0. The force has the same value everywhere on the second leg from (L,0) to (L,L).

Execute

0F dl =rrg

ˆF CLj=r



Example 7.13 (SOLN)

The displacement on this leg is in the +y-direction, so and

Execute

ˆdl dyj=r

ˆ ˆF dl CLj dyj CLdy= =rrg g

The work done on the second leg is then( , ) 2

2 ( ,0) 0 0

L L y L L

L yW F dl CLdy CL dy CL

−

−= = = =∫ ∫ ∫

rrgOn the third leg, from (L,L) to (0,L), is again perpendicular to the displacement so W3 = 0. The force is zero on the final leg, from (0,L) to (0, 0 ), so no work is done and W4 = 0.

Fr



Example 7.13 (SOLN)ExecuteThe work done by the force on the round trip is F

r2 2

1 2 3 4 0 0 0W W W W W CL CL= + + + = + + + =The starting and ending points are the same, but the total work done by is not zero. This is a conservative force; it cannot be represented by a potential energy function.

Fr



Because W is positive, the mechanical energy of the electron increases as it goes around the loop. This is not a mathematical curiosity; it’s a description of what happens in an electrical generating plant. A loop of wire is moved through a magnetic field, which gives rise to a nonconservative force similar to the one in this example. Electrons in the wire gain energy as they move around the loop, and this energy is carried via transmission lines to the customer. (We’ll discuss how this work in detail in Chapter 29.) All the electrical energy used in the home and in industry comes from work done by nonconservative forces!




How would the value of W change if the electron went around the loop clockwise instead of counterclockwise? The force would be unaffected, but the direction of each infinitesimal displacement would reverse. Thus the sign of work would also reverse, and the work for a clockwise round trip would be W = -CL2. This is a different behavior than the nonconservative friction force. When a body slides over a stationary surface with friction, the work done by friction is always negative, no matter what the direction of motion (see Example 7.7 in Section 7.1).


Fr

dlr




The Law of Conservation of Energy

• When a car brakes, the tires and road surface becomes hot. This type of energy that relates with this change in state of the material is called internal energy.

• Raising the temperature will increase the internal energy and vice versa.

• Experiments show that change in internal energy is exactly equal to the absolute value of the work done by friction is . int otherU WΔ =−





• Experiments show that change in internal energy is exactly equal to the absolute value of the work done by friction,

int otherU WΔ =−

where is the change in internal energy.intUΔ• Substitute this into Eq. (7.7),

1 1 int 2 2K U U K U+ −Δ = +

int 0 (7.16)K U UΔ +Δ +Δ =

(law of conservation of energy)





• Law of conservation of energy states that energy is never created or destroy; it only changes form.

• Energy can be convert from one form to another, and at the same time lost in the form of friction and heat energy.



Let’s look again at Example 7.6 in Section 7.1, in which your cousin Throcky skateboards down a curved ramp. He starts with zero kinetic energy and 735J of potential energy, and at the bottom he has 450J of kinetic energy and zero potential energy. So and . The work Wother = Wfric done by the nonconservative friction forces is -285J, so the change in internal energy is .

Example 7.14 Work done by friction

450K JΔ =+735U JΔ =−

int 285otherU W JΔ =− =+



The wheels, the bearings, and the ramp all get a little warmer as Throcky rolls down. In accordance with Eq.(7.16), the sum of the energy changes equals zero:

Example 7.14 Work done by friction

int 450 ( 735 ) 285 0K U U J J JΔ +Δ +Δ =+ + − + =

The total energy of the system (including nonmechanical forms of energy) is conserved.



7.4 Force and Potential Energy

• In physics, there will be situations that you are given an expression for the potential energy as a function of position and have to find the corresponding force.

• Consider a force motion along x-direction by a function and the potential energy as U(x). Recall for any displacement, the work done is , for a small displacement ,

( )xF x

W U=−Δ xΔ( )xF x x UΔ = −Δ

( )xUF xx

Δ= −Δ




• Therefore,( ) ( ) (7.17)x

dU xF x

dx= −

(force from potential energy, one dimension)• This equation tells us that when U increases F also increase, which is great

amount of work is needed for a distance, x, and corresponds to a larger force magnitude.

• Also when is in the positive x-direction, U(x) decreases with increasing x.

( )xF x




• The physical meaning is that a conservative force always acts to push the system toward lower potential energy.

• When substituting into Eq. (7.17),we get the correct expression for the force exerted by an ideal spring,( ) 21

2U x kx=

( ) 212x

dF x kx kxdx⎛ ⎞= − = −⎜ ⎟⎝ ⎠




• We can use the expression to plot graphs of potential energy and force versus position for the spring force.




• For gravitational potential energy, we have

• When substituting into Eq. (7.17), we get ( )U y mgy=

( )y

dU d mgyF mgdy dy

=− =− =−

• This is the correct expression for gravitational force.




• We can use the expression to plot graphs of potential energy and force versus position for the gravitational force.



An electrically charged particle is held at rest at the point x = 0, while a second particle with equal charge is free to move along the positive x-axis. The potential energy of the system is

Example 7.15 An electric force and its potential energy

( ) CU xx

=

Where C is a positive constant that depends on the magnitude of the charges. Derive an expression for the x-component of force acting on the movable charge, as a function of its position.



We are given the potential-energy function U(x). Hence we can use Eq.(7.17) to find our target, the function Fx(x).

Example 7.15 (SOLN)

The derivative with respect to x of the function 1/x is -1/x2. So the force on the movable charge for x > 0 is

Identify and Set up

Execute

2 2( ) 1( )x

dU x CF x Cdx x x

⎛ ⎞=− =− − =⎜ ⎟⎝ ⎠



The x-component of force is positive, corresponding to a repulsive interaction between like electric charges. The potential energy is very large for small x and approaches zero as x becomes large; the force pushes the movable charge towards large positive values of x, for which the potential energy is less. The force varies as 1/x2; it is small when the particles are far apart (large x) but becomes large when the particles are close together (small x). This is an example of Coulomb’s law for electric interactions, which we will study in greater detail in Chapter 21.





Force and Potential Energy in Three Dimensions

• A force can have 3 components and each components may be a function of x, y and z.

• Use Eq. (7.17) to find each component,

( ), and x y zF F F

x y zU U UF F Fx y z

Δ Δ Δ=− =− =−Δ Δ Δ

• We compute the derivative of U with respect to x, y and z,

(7.18)x y zU U UF F Fx y z

∂ ∂ ∂=− =− =−∂ ∂ ∂

(force from potential energy)





• Writing in unit vectors, the expression is

$ $ (7.19)U U UF i j kx y z

∂ ∂ ∂⎛ ⎞=− + +⎜ ⎟∂ ∂ ∂⎝ ⎠ur $

(force from potential energy)

• The expression can further reduced with a an operation called the gradient of U, thus the force is the negative of the gradient of the potential-energy function:

(7.20)F U=−∇ur ur





• As a check, substitute Eq. (7.20) into the function U = mgy for gravitational potential energy:

• This is the same as the expression for gravitational force.

( ) ( ) ( )$ ( )$ ( )$mgy mgy mgyF mgy i j k mg j

x y z∂ ∂ ∂⎛ ⎞= −∇ = − + + = −⎜ ⎟∂ ∂ ∂⎝ ⎠

ur ur $



A puck slides on a level, frictionless air-hockey table. The coordinates of the puck are x and y. It is acted on by a conservative force described by the potential-energy function

Example 7.16 Force and potential energy in two dimensions

2 21( , ) ( )2

U x y k x y= +

Derive an expression for the force acting on the puck, and find an expression for the magnitude of the force as a function of position.



We’ll find the components of the force from the function U(x,y) using Eq.(7.18), then determine the magnitude of the force using the formula for the magnitude of a vector:


2 2x yF F F= +

The components of the force areExecute

xUF kxx

∂=− =−

∂ yUF kyy

∂=− =−

∂



From Eq.(7.19) this corresponds to the vector expression


ˆ ˆ( )F k xi yj=− +r

Now is just the position vector of the particle, so we can rewrite this expression as . This represents a force that at each point is opposite in direction to the position vector of the point – that is, a force that at each point is directed toward the origin. The potential energy is minimum at the origin, so again the force pushes in the direction of decreasing potential energy.

ˆ ˆxi yj+ rrF kr=−r r



The magnitude of the force at any point is


Where r is the particle’s distance from the origin. This is the force exerted by a spring that obeys Hooke’s law and has a negligibly small length (compared to the other distances in the problem) when it is not stretched. So the motion of the puck is the same as if it were attached to one end of an ideal spring of negligible unstretched length; the other end is attached to the air-hockey table at the origin.

2 2 2 2( ) ( )F kx ky k x y kr= − + − = + =



We can check our result by noting that the potential-energy function can also be expressed as U = ½ kr2. Written this way. U is a function of a single coordinate r, so we can find the force using Eq.(7.17) with x replaced by r:


212r

dU dF kr krdr dr

⎛ ⎞=− =− =−⎜ ⎟⎝ ⎠Just as we calculated above, the force has magnitude kr; the minus sign indicates that the force is radially inward (toward the origin).



7.5 Energy Diagrams

• By looking at the graph of the potential-energy function U(x), we are able to understand the motions when a particle moves along a straight path.

• The figure shows a glider with mass m that moves along the x-direction.



7.5 Energy Diagrams

• Below is the graph of the corresponding potential-energy function . The limits of the motion are the points where the U curve intersects the horizontal line representing the total mechanical energy E.

( ) 212

U x kx=



7.5 Energy Diagrams

• K can never be negative and the potential energy U can never be greater than the total energy E.• When x = 0, the slope and the force is zero.• When x is positive, the slope is positive and the force is negative.• When x is negative, the slope is negative and the force is positive.• We say that x = 0 is a point of stable equilibrium, more generally, any minimum in a potential-energy

curve is a stable equilibrium position.

xF

xF



7.5 Energy Diagrams



7.5 Energy Diagrams

• Previously graph (a) below shows a more general potential-energy function U(x) while graph (b) shows the corresponding force .

• is zero at point and . They are the stable equilibrium points. The slope of U(x) is zero at these points.

• When the particle is displaced to the other sides, the force pushes back towards the equilibrium point.

/xF dU dx=−

xF 1x 3x



• The slope is zero at points and .• However when the particle is displaced a little to the right of either sides, slope pf U(x) is negative. • Therefore these points are called unstable equilibrium points; any maximum in a potential-energy

curve is an unstable equilibrium position. • Again, U cannot be greater than because K can’t be negative. • and are the turning point of the particle’s motion as it stops and reverse its direction.

7.5 Energy Diagrams

2x 4x

1E

ax bx



7.5 Energy Diagrams

• When increase potential energy to , particle can move from to .• When energy greater than , the particle moves to indefinitely large

values of x.• represents the least possible total energy the system can have.

2Ecx dx

3E

0E

• Direction of the body is determined by the sign of potential energy U.• is the physical significant quantity difference in the value

of U between 2 points.

/xF dU dx=−



• The total potential energy is the sum of the gravitational and elastic energy. If no forces other than gravitational and elastic forces do work on a particle, the sum of kinetic and potential energy is conserved. This sum E = K + U is called the total mechanical energy.

• When forces other than gravitational and elastic forces do work on a particle, the work done by these other forces equals the change in the total mechanical energy (kinetic plus potential energy).

Concept Summary

otherW



• All forces are either conservative or nonconservative. A conservative force is one for which the work-kinetic energy relation is completely reversible. The work of a conservative force can always be represented by a potential-energy function, but the work of a non-conservative force cannot.

• The work done by nonconservative forces manifests itself as change in the internal energy of bodies. The sun of kinetic, potential and internal energy is always conserved.

Concept Summary



• For motion along a straight line, a conservative force is the negative derivative of its associated potential-energy function U. In three dimensions, the components of a conservative force are negative partial derivative of U.

Concept Summary

( )xF x



Key Equations

1 2gravW mgy mgy= −

1 2 (7.1), (7.3)U U U= − =−Δ

2 21 2

1 12 2elW kx kx= −

1 2 (7.10)U U U= − =−Δ

1 1 2 2 (7.11)K U K U+ = +



Key Equations

1 1 2 2 (7.13)otherK U W K U+ + = +

int 0 (7.16)K U UΔ +Δ +Δ =

( ) ( ) (7.17)xdU x

F xdx

= −

, , (7.18)x y zU U UF F Fx y z

∂ ∂ ∂=− =− =−∂ ∂ ∂

$ $ (7.19)U U UF i j kx y z

∂ ∂ ∂⎛ ⎞=− + +⎜ ⎟∂ ∂ ∂⎝ ⎠ur $

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Module 7 Work, Energy, Power