Work done on an object = energy gained by ... - HHS Physics

Energy Intro.notebook

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February 13, 2013

Energy

Energy can be defined as the capacity for doing work, or the property of a system that diminishes when the system does work on any other system by an amount equal to the work done.

1) When work is done on an object, the energy of that object increases.

2) When the object does work on something else, it loses energy.

Example: A weightlifter lifts a barbell from the ground to over her head. The weight lifter applied a force to the barbell which caused the barbell to move, thus doing work on the barbell. The GPE of the barbell increased, the energy possessed by the weightlifter decreased. The amount of energy gained by the barbell is equal to the amount of energy lost by the weightlifter.

Work was done ON the barbell

Work was done BY the weightlifter

Work done on an object = energy gained by the object

Work done by an object = energy lost by the object


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February 13, 2013

Forms of Energy

Kinetic Energy (KE) Is the energy possessed by an object because it is in motion. It is sometimes referred to as the energy of motion. The transitional KE of an object is a property of the objects mass and velocity. The formula for determining KE is:

KE = 1/2 m V2

Gravitational Potential Energy (Ug or GPE) is the energy possessed by an object due to its position in a gravitational field. It is sometimes referred to as the energy of position. This vertical distance is measured relative to an arbitrary zero level, often the surface of earth. The Gravitational Potential Energy is a function of the mass of the object and the height of the object above the zero (or reference) level in a gravitational field. The formula for determining GPE is:

GPE = mgh


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February 13, 2013

Forms of Energy

Elastic Potential Energy (Ue or EPE) is Potential energy stored as a result of deformation of an elastic object, such as the stretching of a spring. It is equal to the work done to stretch the spring, which depends upon the spring constant k as well as the distance stretched. According to Hooke's law, the force required to stretch the spring will be directly proportional to the amount of stretch. Since the force has the form F = kx, then the work done to stretch the spring a distance x is

EPE = 1/2 k x2

Internal energy (U) is defined as the energy associated with the random, disordered motion of molecules. It is separated in scale from the macroscopic ordered energy associated with moving objects; it refers to the invisible microscopic energy on the atomic and molecular scale. The internal energy (U) is the sum of all forms of energy (Ei) intrinsic to a thermodynamic system:

U = ΣEi

Note: Internal energy is NOT a form of mechanical energy.


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February 13, 2013

Mechanical energy is the sum of potential energy and kinetic energy present in the components of a mechanical system. It is the energy associated with the motion and position of an object. The law of conservation of energy states that in an isolated system that is only subject to conservative forces, like the gravitational force, the mechanical energy is constant. If an object is moved in the opposite direction of a conservative net force, the potential energy will increase and if the speed (not the direction of velocity but its magnitude) of the object is changed, the kinetic energy of the object is changed as well.

A conservative force is a force with the property that the work done in moving a particle between two points is independent of the path taken. Equivalently, if a particle travels in a closed loop, the net work done (the sum of the force acting along the path multiplied by the distance travelled) by a conservative force is zero.

A conservative force is dependent only on the position of the object. If a force is conservative, it is possible to assign a numerical value for the potential energy at any point. When an object moves from one location to another, the force changes the potential energy of the object by an amount that does not depend on the path taken. If the force is not conservative, then defining a scalar potential is not possible, because taking different paths would lead to conflicting potential differences between the start and end points.

Gravity is an example of a conservative force, while friction is an example of a nonconservative force.


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February 13, 2013

Mechanical Energy (ME) is equal to the Kinetic Energy (KE) plus the Gravitational Potential Energy (GPE) plus the Elastic Potential Energy (EPE)

Note: Potential Energy often is given the variable U, with Gravitational Potential Energy is Ug and Elastic Potential Energy is Ue.

ME = KE + GPE + EPE

ME = KE + Ug + UE

In a conservative system the Mechanical energy at any point is the same!

Mechanical energy must be conserved under conservative forces, but the kinetic energy can fluctuate based on the speed of the particles in the system, there must be an additional quantity of energy that is a property of the structure of the system. This quantity, potential energy, is denoted by the symbol U and can be easily derived from our knowledge of conservative systems.

Consider a system under the action of a conservative force. When work is done on the system it must in some way change the velocity of its constituent parts (by the Work Energy Theorem), and thus change the configuration of the system. We define potential energy as the energy of configuration of a conservative system, and relate it to work in the following way:

∆U = Work


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February 13, 2013

If a 2 kg ball is dropped 10 m from rest and there is no friction the ball will obey this conservation principle.

h = 0 m

hA = 10 m

VA = 0Since VA = 0 and there are no springs KEA = 0 and EPEA = 0. So,

MEA = KEA + GPEA + EPEA

MEA = GPEA = mghAMEA = (2 kg)(10 m/s2)(10 m) = 200 J

A MEA = KEA + GPEA + EPEA

0 0


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February 13, 2013

hB = 5 m

VB = 10 m/s

MEB = KEB + GPEB + EPEB

Since there are no springs EPEB = 0. So,

MEB = KEB + GPEB = 1/2 mVB2 + mghB

MEB = 1/2 (2 Kg)(10 m/s)2 + (2 kg)(10 m/s2)(5 m)

MEB = 100 J + 100 J = 200 J

B

After the ball has fallen 5 m, using kinematics we find it has a velocity of 10 m/s. What is its ME ?

h = 0 m


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February 13, 2013

hC = 3 m

VC = 11.832 m/s

MEC = KEC + GPEC + EPEC

Since there are no springs EPEC = 0. So,

MEC = KEC + GPEC = 1/2 mVC2 + mghC

MEC = 1/2 (2 Kg)(11.832 m/s)2 + (2 kg)(10 m/s2)(3 m)

MEC = 140 J + 60 J = 200 JC

After the ball has fallen 7 m, using kinematics we find it has a velocity of 11.832 m/s. What is its ME ?

h = 0 m


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February 13, 2013

hD = 0 m

VD = 14.14 m/s

MED = KED + GPED + EPED

Since there are no springs EPEC = 0, and hD =0, so there is no GPE. So,

MED = KED = 1/2 mVD2

MED = 1/2 (2 Kg)(14.14 m/s)2

MED = 200 J

D

After the ball has fallen 10 m, using kinematics we find it has a velocity of 14.14 m/s. What is its ME ?

h = 0 m


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February 13, 2013

A direct consequence of the closed path test is that the work done by a conservative force on a particle moving between any two points does not depend on the path taken by the particle. Also the work done by a conservative force is equal to the negative of change in potential energy during that process. For a proof of that, let's imagine two paths 1 and 2, both going from point A to point B. The variation of energy for the particle, taking path 1 from A to B and then path 2 backwards from B to A, is 0; thus, the work is the same in path 1 and 2, i.e., the work is independent of the path followed, as long as it goes from A to B.

For example, if a child slides down a frictionless slide, the work done by the gravitational force on the child from the top of the slide to the bottom will be the same no matter what the shape of the slide; it can be straight or it can be a spiral. The amount of work done only depends on the vertical displacement of the child.

∆H


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February 13, 2013

1) Consider the falling and rolling motion of the ball in the following two resistancefree situations. In one situation, the ball falls off the top of the platform to the floor. In the other situation, the ball rolls from the top of the platform along the staircaselike pathway to the floor. For each situation, indicate what types of forces are doing work upon the ball. Indicate whether the energy of the ball is conserved and explain why. Finally, fill in the blanks for the 2kg ball.

2) If frictional forces and air resistance were acting upon the falling ball in #1 would the kinetic energy of the ball just prior to striking the ground be more, less, or equal to the value predicted in #1?


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February 13, 2013

Use the following diagram to answer the questions below. At Point A the ball has a velocity of 2 m/s to the right. Neglect the affect of resistance forces.

1) What forces are acting on the ball at each point?A) Gravity, Normal, and Friction B) Gravity, Normal, and Compression C) Gravity, Normal D) Normal, and Friction E) Gravity, and Friction F) Compression, and Gravity

2) As the object moves from point A to point D across the surface, the sum of its gravitational potential and kinetic energies ____.

A) decreases, only B) decreases and then increases C) increases and then decreases D) remains the same

3) The object will have a minimum gravitational potential energy at point ____.4) The object's kinetic energy at point C is less than its kinetic energy at point ____.


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February 13, 2013

Find The Quantities at each point in the diagram above. Neglect the affect of resistance forces. Mass = 1 Kg


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February 13, 2013

Find the velocity at each point in the diagram above. M = 100 kg. Neglect the affect of resistance forces.


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February 13, 2013

A ball of mass m falls past a window with a vertical height of h and with no resistive forces acting on it. The velocity of the ball at the top of the window is VA towards the ground. Derive an equation to find the velocity at the bottom of the window, VB.

h

A

B


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February 13, 2013

A ball of mass m falls past a window with a vertical height of h and with no resistive forces acting on it. The ball at the top of the window has a velocity of VA towards the ground and at the bottom of the window has a velocity of VB towards the ground. Derive an equation to solve for the height of the window h.

h

A

B


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February 13, 2013

The Spring Force

The force exerted by a spring on a mass m can be calculated using Hooke's law. In physics, Hooke's law of elasticity is an approximation that states that the amount by which a material body is deformed (the strain) is linearly related to the force causing the deformation (the stress). Materials for which Hooke's law is a useful approximation are known as linearelastic or "Hookean" materials.

For systems that obey Hooke's law, the extension produced is proportional to the load:

Fspring = k x

where

1) x is the distance the spring is elongated or compressed by

2) F is the restoring force exerted by the spring

3) k is the spring constant or force constant of the spring

When this holds, we say that the spring is a linear spring.

The point at which a material ceases to obey Hooke's Law is known as its elastic limit.


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February 13, 2013

+

∆X1 = 1∆X2 = 2

FSFS


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February 13, 2013

Elastic Potential Energy

Springs are normally assumed to be massless so their inertia can be neglected. This also means that the force exerted by both ends of the spring are the same but in opposite directions.

The potential energy stored in a spring is given by

Ue = EPE = 1/2 k x2

which comes from adding up the energy it takes to incrementally compress the spring. That is, the integral of force over distance. (Note that potential energy of a spring is always positive.)

This potential can be visualized as a parabola on the U vs. X graph. As the spring is stretched in the positive xdirection, the potential energy increases (the same thing happens as the spring is compressed). The corresponding point on the potential energy curve is higher than that corresponding to the equilibrium position (x=0). The tendency for the spring is to therefore decrease its potential energy by returning to its equilibrium (unstretched) position, just as a ball rolls downhill to decrease its gravitational potential energy.


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February 13, 2013

What is the total energy of the massspring system shown below? The mass is shown at its maximum displacement on the spring, 5 meters from the equilibrium point.

h = 0

(at rest)


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February 13, 2013

You are designing a game and want to see if the ball’s maximum speed is sufficient to require the use of goggles. In your game, a 0.015 kg ball is to be shot from a spring gun whose spring has a spring constant of 600 N/m. The spring will be compressed 0.05 m when in use.

A) How fast will the ball be moving as it leaves the gun?

B) how high will the ball go if the gun is aimed vertically upward?

X = 0.05 m


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February 13, 2013

The spring in the muzzle of a child's spring gun has a spring constant of 780 N/m. To shoot a ball from the gun, first the spring is compressed and then the ball is placed on it. The gun's trigger then releases the spring, which pushes the ball through the muzzle. The ball leaves the spring just as it leaves the outer end of the muzzle. When the gun is inclined upward by 90° to the horizontal, a 0.059 kg ball is shot to a maximum height of 1.75 m. Assume air drag on the ball is negligible.

A) Assuming that friction on the ball within the gun can be neglected, find the spring's initial compression distance (X).

B) At what speed does the spring launch the ball?

X

1.75 M

h = 0


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February 13, 2013

A 20 kg mass, released from rest, slides 6 meters down a frictionless plane inclined at an angle of 30º with the horizontal and strikes a spring of spring constant K = 200 newtons/meter as shown in the diagram above. Assume that the spring is ideal, that the mass of the spring is negligible, and that mechanical energy is conserved.

a. Determine the speed of the block just before it hits the spring.

b. Determine the distance the spring has been compressed when the block comes to rest.

c. Is the speed of the block a maximum at the instant the block strikes the spring?

hC

∆H6 m

30o

hA

hB


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February 13, 2013

A .5 kg ball is released from rest with a vertical height of 1.25 m above the floor. It then rolls along a horizontal floor before hitting a stop block that is connected to a spring. The spring is depressed 4 cm before stopping the ball. If all resistive forces can be ignored, find the spring constant K for the spring.

V = 0

X = 0.04 m


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February 13, 2013

Energy Transformation for a Pendulum

The motion of a pendulum is a classic example of mechanical energy conservation. A pendulum consists of a mass (known as a bob) attached by a string to a pivot point. As the pendulum moves it sweeps out a circular arc, moving back and forth in a periodic fashion. Neglecting air resistance (which would indeed be small for an aerodynamically shaped bob), there are only two forces acting upon the pendulum bob. One force is gravity; the force of gravity acts in a towards the center of Earth and does work upon the pendulum bob. However, gravity is an internal force (or conservative force) and thus does not serve to change the total amount of mechanical energy of the bob. The other force acting upon the bob is the force of tension. Tension is an external force and if it did do work upon the pendulum bob it would indeed serve to change the total mechanical energy of the bob. The force of tension does not do work since it always acts in a direction perpendicular to the motion of the bob. At all points in the trajectory of the pendulum bob, the angle between the force of tension and its instantaneous displacement is 90 degrees. Thus, the force of tension does not do work upon the bob [WFt = Ft S cos(90o) is 0 Joules].


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February 13, 2013

A simple pendulum is hung form a point O. The bob of the pendulum with mass m, is stationary at point A. It is then set into motion from one extreme position B. The bob swings from one extreme position B to another extreme position C via position A.

You can easily observe that

1. at position A : the GPE = 0

2. at position B : the GPE = maximum = mgh, the bob stops for a fraction of a second, hence its

KE = 0

3. at position C : the GPE = maximum = mgh the bob stops for a fraction of a second, hence its

KE = 0

4. at position A, notice that the velocity of the swing is high, Thus at A the KE = maximum.

Thus as the bob swings back and forth, the KE and the GPE vary from 0 to maximum, but in the opposite direction. That is, as the KE changes from 0 to maximum, the GPE changes from maximum to 0. Thus the sum of the KE + GPE remains constant. Mechanical Energy is conserved!


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February 13, 2013

A 2 kg pendulum bob is hung on a 1 m long chord and is raised to an angle of 60o as seen in the figure to the left.

What is the total Mechanical Energy at Point A?

600L = 1 m

h = 0

hA

MEA = KEA + GPEA + EPEA

Since there are no springs EPE = 0, and because VA = 0; KEA = 0: Thus MEA = GPEA

MEA = GPEA

MEA = mghA

So, we need to calculate hA. This is done using trigonometry. Notice the pendulum forms a triangle at its start position.


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February 13, 2013

600L = 1 m

h = 0

hA

600 L = 1 mH

We can find H using the cosine function

cos θ = Adjacent side divided by the hypotenuse

θ = 60o, Adjacent side = H, and hypotenuse = 1 m

cos 60o = H

1 mH = (1 m) cos 60o = 0.5 m

H

H is NOT the height above zero we need to calculate GPE, but it is needed because H + hA = length of the pendulum.

hA


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February 13, 2013

Internal vs. External Forces

There are a variety of ways to categorize all the types of forces. We know that all the types of forces can be categorized as contact forces or as actionatadistance forces. Whether a force was categorized as an actionatadistance force was dependent upon whether or not that type of force could exist even when the objects were not physically touching. The force of gravity, electrical forces, and magnetic forces were examples of forces that could exist between two objects even when they are not physically touching. Now, we will learn how to categorize forces based upon whether or not their presence is capable of changing an object's total mechanical energy. We will learn that there are certain types of forces, that when present and when involved in doing work on objects will change the total mechanical energy of the object. And there are other types of forces that can never change the total mechanical energy of an object, but rather can only transform the energy of an object from potential energy to kinetic energy (or vice versa). The two categories of forces are referred to as internal forces and external forces.


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February 13, 2013

Internal Forces are forces between objects found inside the system.

cannot do work on systemcannot change total energy of the system

External forces are forces that act on the system and their agents are part of the environment.

can do work on system

can transfer energy into or out of system, thus changing the total energy of the system

For our purposes, we will simply say that external forces include the applied force, compression force, tension force, friction force, and air resistance force. And for our purposes, the internal forces include the gravity forces, magnetic force, electrical force, and spring force.

While this is a simplistic approach, it is an approach that will serve us well in our introduction to physics.


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February 13, 2013

Analysis of Situations Involving External Forces

Whenever work is done upon an object by an external force, there will be a change in the total mechanical energy of the object. If only internal forces are doing work (no work done by external forces), there is no change in total mechanical energy; the total mechanical energy is said to be conserved. Because external forces are capable of changing the total mechanical energy of an object, they are sometimes referred to as nonconservative forces. Because internal forces do not change the total mechanical energy of an object, they are sometimes referred to as conservative forces.

The quantitative relationship between work and mechanical energy is expressed by the following equation:

MEI + Wext = MEF


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February 13, 2013

The equation states that the initial amount of total mechanical energy (MEI) plus the work done by external forces (Wext) is equal to the final amount of total mechanical energy (MEF). A few notes should be made about the equation. First, the mechanical energy can be either potential energy (in which case it could be due to springs or gravity) or kinetic energy. Given this fact, the equation can be rewritten as

KEI + GPEI + EPEI + Wext = KEF + GPEF + EPEF

Wext = ∆KE + ∆GPE + ∆EPE

Fext s cosθ = 1/2 m(VF2 VI2) + mg(hF hI) + 1/2 k(XF2 XI2)

The second note that should be made about the above equation is that the work done by external forces can be a positive or a negative work term. Whether the work term takes on a positive or a negative value is dependent upon the angle between the force and the motion. Recall that the work is dependent upon the angle between the force and the displacement vectors. If the angle is 180 degrees as it occasionally is, then the work term will be negative. If the angle is 0 degrees, then the work term will be positive.


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February 13, 2013

Now we will investigate situations involving work being done by external forces (nonconservative forces). Consider a weightlifter who applies an upwards force (say 1000 N) to a barbell to displace it upwards a given distance (say 0.25 meters) at a constant speed. The initial energy plus the work done by the external force equals the final energy. If the barbell begins with 1500 Joules of energy (this is just a made up value) and the weightlifter does 250 Joules of work (Fscos θ = 1000 N 0.25 m cosine 0 degrees = 250 J), then the barbell will finish with 1750 Joules of mechanical energy. The final amount of mechanical energy (1750 J) is equal to the initial amount of mechanical energy (1500 J) plus the work done by external forces (250 J).

s


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February 13, 2013

A 1000kg car traveling with a speed of 25 m/s skids to a stop. The car experiences an 8000 N force of friction. Determine the stopping distance of the car.


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February 13, 2013

At the end of the Bizarro roller coaster ride, the 6000kg train of cars (includes passengers) is slowed from a speed of 25 m/s to a speed of 4 m/s over a distance of 22 meters. Determine the braking force required to slow the train of cars by this amount.

VF = 4 m/sVI = 25 m/s

22 m


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February 13, 2013

1) What force is required to stop the car? Convert the answer from newtons to pounds (To convert N to tons multiply by force by 0.00011 ton/N).

A 1275.5 kg car traveling on a horizontal road with a speed of 17.8816 m/s (40 mph) crashes into a tree. The car collapses a distance of .4025 m during the crash.


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February 13, 2013

2) Luckily, the 72.7 kg driver was wearing his seatbelt. If the seat belt harness stretches, increasing stopping distance by 50%: 0.60375 m, What force is required to stop the driver? (To convert N to tons multiply by force by 0.00011 ton/N).


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February 13, 2013

3) Assume the 72.7 kg driver was not wearing his/her seatbelt (An AP Bio Student). If not wearing seatbelt, stopping distance determined by nature of collision with windshield, steering column, etc.: stopping distance 0.061 m, What force is required to stop the driver? Convert your answer from newtons to pounds (To convert N to tons multiply by force by 0.00011 ton/N).


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February 13, 2013

A rope is attached to a 50.0kg crate to pull it up a frictionless incline to a height of 3meters. A diagram of the situation and a freebody diagram are shown below. The crate is initially at rest and is at rest after it reaches the top of the incline. Note that the force of gravity has two components (parallel and perpendicular component); the parallel component balances the applied force and the perpendicular component balances the normal force.

1) Of the forces acting upon the crate, which one(s) do work upon it?

2) Based upon the types of forces acting upon the system and their classification as internal or external forces, is energy conserved? Explain.

3) Calculate the amount of external work done upon the crate.

Ft


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February 13, 2013

Work and Kinetic Energy

When a net force does work on a rigid body, it causes the body's speed to change.

The work done by the net force is the same as the sumtotal of the work done by the action of every force acting the body. If you add up the work done by each of the forces acting on a body you will get the same value as the work done by the net force.

The work done on a body that caused the body to be set in motion with some speed V can be expressed as function of the body's final speed v and mass m, independent of type of force that acted on the body. We call this function the body's Kinetic Energy.


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February 13, 2013

WorkEnergy Theorem

The energy associated with the work done by the net force does not disappear after the net force is removed (or becomes zero), it is transformed into the Kinetic Energy of the body. We call this the WorkEnergy Theorem.

WNET = ∆KE = KEF KEI

WNET = 1/2 m(VF2 VI

2)

If the body's speed increases, then the work done on the body is positive and we say its Kinetic Energy has increased. Whereas if the body's speed decreases then it kinetic energy decreases and the change in kinetic energy ∆KE is negative. In this case the body does positive work on the system slowing it down or alternately the work done on the body is negative.

If the object is not rigid and any of the forces acting on it deforms the object, then the WorkEnergy Theorem will no longer be valid. Some of the energy transferred to the object has gone into deforming the object and is no longer available to increase or decrease the object's Kinetic Energy.


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February 13, 2013

KEI = 1/2 mVI2

S

VIFNETVF

KEF = 1/2 mVF2

WNET > 0∆V > 0KEF > KEI

WNET = FNET s cos (0o) = + FNET

KEI = 1/2 mVI2

S

VIFNET VF

KEF = 1/2 mVF2

WNET < 0∆V < 0KEF < KEI

WNET = FNET s cos (180o) = FNET


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February 13, 2013

S

VIFNETVF

Using Newtons Second Law: Fnet = ma

Find A using equation 3 of the Big Three:

VF2 = VI

2 + 2aDF

DF = s

so

VF2 VI

2 = 2a s

Solving for a s

a s = VF2 VI

2

2

Wnet = Fnetscosθ

θ = 0o

Wnet = Fnet s

Wnet = ma s

Substitute for a s

Wnet = m (VF2 VI

2)2

Wnet = 1/2 m (VF2 VI

2)

but

∆KE = 1/2 m (VF2 VI

2)

so

Wnet = ∆KE


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February 13, 2013

Calculus derivation of the Work Energy Theorem

The energy associated with the work done by the net force does not disappear after the net force is removed (or becomes zero), it is transformed into the Kinetic Energy of the body. We call this the WorkEnergy Theorem.

To reduce the complexity of the derivation, we will assume that the direction of the Net Force is constant while the work is being done. The WorkEnergy Theorem is still valid if the net force changes direction as well as magnitude while the work is being done, provided the body is rigid.

S

VIFNETVF


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February 13, 2013

The key step is to convert the calculus definition for acceleration into an expression that is a derivative of x. We know Acceleration is the first derivative of velocity and the velocty is the first derivative of position.

A = dVdt

V =dSdt

Plug this and the Second Law into the definition for Work, and integrate.

Wnet = ∫ Fnet dS = ∫ mA dS = ∫ m dS

Wnet = ∫ m Vdt

Wnet = ∫ m V dV

Wnet = ½ m V2

Evalutated from initial to final

Wnet = ½ m VF2 ½ m VI

2

Wnet = KEF KEI = Δ KE

Wnet = Δ KE

V dt = dS

dVdtdV

dt


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February 13, 2013

m = 5 kgµk = .2 FT = 100 N

FT = 100 N

FN = 50 N

Fg = 50 N

f = 10 N S5 m

A 5 kg block is slid across a horizontal surface by a 100 N tension force. If the block was initially moving with a constant velocity of 5 m/s to the right, how fast is it moving after it has slid 5 m?

1) Find the work done by all the forces.2) Find the velocity of the block after 5 s using MEI + Wext = MEF where Wext is the all the non zero works calculated in 1. MEI + WFt + Wf = MEF3) Find the velocity using the workenergy theorem.

Fnet = 90 N at 0o


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February 13, 2013

A 1250 kg car that has a 300 hp engine can accelerate from 0 to 60 mph (26.8224 m/s) in 6 s along a horizontal road. The car will travel 80.46 m in that time. Calculate the net force that the car needs to accelerate in this way.

80.46 m


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February 13, 2013

A 1250 kg car that has a 300 hp engine can accelerate from 0 to some VF along a horizontal road. The car will travel 40 m in that time. The net force that the car experiences is 5500 N. Calculate the final velocity of the car.

40 m


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February 13, 2013

A 1250 kg car that has a 300 hp engine can cruise at 26.8224 m/s along a horizontal road. If he car travels 100 m in that time. Calculate the work that the cars engine does to keep the car moving at this constant velocity, if the drag force is 54707.386 N.

100 m


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February 13, 2013

Power

Power is the transfer rate of energy or work done per unit of time. It is given the symbol: P. Thus, power follows the rate equation:

Average Power = Work done by a force / Time taken to do this work

Pavg = W / ∆t

The unit of power is the joule per second (J/s), known as the Watt (W) (in honor of James Watt, the eighteenthcentury developer of the steam engine).

For example, the rate at which a light bulb transforms electrical energy into thermal energy and light is measured in watts, the more wattage, the more power, or equivalently the more electrical energy is used per unit time.

Another common unit is the horsepower (hp); 1 hp = 746 W.



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February 13, 2013

Energy transfer can be used to do work, so power is also the rate at which this work is performed. The same amount of work is done when carrying a load up a flight of stairs whether the person carrying it walks or runs, but more power is expended during the running because the work is done in a shorter amount of time. The output power of an electric motor is the product of the torque the motor generates and the angular velocity of its output shaft. The power expended to move a vehicle is the product of the traction force of the wheels and the velocity of the vehicle.

Most machines are designed and built to do work on objects. All machines are typically described by a power rating. The power rating indicates the rate at which that machine can do work upon other objects. Thus, the power of a machine is the work/time ratio for that particular machine. A car engine is an example of a machine that is given a power rating. The power rating relates to how rapidly the car can accelerate the car. Suppose that a 40horsepower engine could accelerate the car from 0 mph to 60 mph in 16 seconds. If this were the case, then a car with four times the horsepower could do the same amount of work in onefourth the time. That is, a 160horsepower engine could accelerate the same car from 0 mph to 60 mph in 4 seconds. The point is that for the same amount of work, power and time are inversely proportional. The power equation suggests that a more powerful engine can do the same amount of work in less time.


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February 13, 2013

The expression for power is work/time. And since the expression for work is force times displacement when θ equals zero, the expression for power can be rewritten as (force times displacement)/time. Since the expression for velocity is displacement/time, the expression for power can be rewritten once more as force times velocity. This is shown below.

PAVG = F v


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February 13, 2013

An escalator is used to move 20 passengers every minute (60 s) from the first floor of a department store to the second. The second floor is located 5.20 meters above the first floor. The average passenger's mass is 54.9 kg. Determine the power requirement of the escalator (units of BOTH Watts and Horsepower) in order to move this number of passengers in this amount of time.

5.20 m

Constant Velocity


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February 13, 2013

A motorized winch is rated at 10.0 kW. At what maximum constant velocity can this winch raise a mass of 27,500 kg?

27,500 kg

Winch


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February 13, 2013

A 1200 kg car can accelerate from rest to a speed of 17.8816 m/s in 4.09 s along a horizontal road. (neglect friction) What is the minimum average horsepower that the engine must deliver to do this?

VI = 0 VF = 17.8816 m/s


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February 13, 2013

A 1100 kg car that has an engine that can deliver 175 hp. What is the cars maximum speed after 10 s if the car starts from rest and is on a horizontal surface? (friction can be neglected)

VI = 0 VF = ?


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February 13, 2013

A 1100 kg car accelerates from 10 m/s to 25 m/s in 8 s while going up a 9 degree inclined plane. (neglect friction) What is the minimum horsepower that the engine must deliver to do this?

9ohA = 0

H

hB = HS


Wext = 1/2 m(VB2 VA

2) + mg(hB hA)P =

Wext

∆t

0

P =1/2 m(VB

2 VA2) + mg(hB hA)

∆t


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February 13, 2013

Water flows from a reservoir at the rate of 4500 kg/min to a turbine 120 m below. Compute the horsepower output of the turbine. (neglect friction in the pipe and the small KE of the water leaving the turbine)

P =Wext

∆t


Wext = mg(hB hA)

00

P =mg(hB hA)

∆t

P =m∆t

g(hB hA)

m∆t

is a mass flow rate in (kg/s)


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February 13, 2013

In unloading the grain from the hold of a ship, an elevator lifts the grain a distance of 15 m. The grain is discharged from the top of the elevator at a rate of 2.5 kg/s, and the discharge speed of each grain particle is 3.5 m/s. Find the minimum horsepower motor that can elevate grain this way?

P =Wext

∆t


Wext = 1/2 m(VB2 VA

2) + mg(hB hA)

0


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February 13, 2013

A winch is used to slide a 325 kg load, at a constant speed, 3.25 m down a 27.0° incline. The rope between the winch and the load is parallel to the incline. The coefficient of kinetic friction between the ramp and the load is 0.40 (f = 1158.3 N). Calculate the force exerted by the winch, the work done by the winch on the load, the work done by the friction force, the work done by the force of gravity, and the net work done on the load. If it takes 60 s to lower the load, calculate the power of the winch.

FN

f = 1158.3 N

Fg = 3250 NX = 1475.5 NY = 2895.8 N

Ft = 317.2 NS

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February 13, 2013

A Tbar tow is planned in a new ski area. At any one time, it will be required, to pull a maximum of 80 skiers up a 600 m slope inclined 15° above the horizontal at a speed of 2.50 m/s. The coefficient of kinetic friction between the skiers skis and the snow is typically 0.055 (f = 3.93 N per skier). As the manager of the facility, what motor power should you request of the construction contractor if the mass of the average skier is 74.0 kg. Assume you want to be ready for any emergency and will order a motor whose power rating is 50 percent larger than the bare minimum.

FN

f

Fg

Ft

S

75


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February 13, 2013

Potential Energy vs Position Graph

Potential Energy = U1. Total Mechanical Energy U = kinetic energy (since KE + U = ME) 2. Where U = ME, kinetic energy is zero. Object is at a momentary stop.

U is at its maximum value. 3. Where U < ME, kinetic energy is positive. Object is moving. 4. Where U > ME, forbidden region. Object cannot be there without

being given more energy. • Slope = opposite of force (F = dU/dx) The force is the negative of the

slope (derivative) on the potential energy versus position graph.• Sloping up (+ such as ) = force is to left or toward smaller values of

position. > example: gravitational potential energy is sloping up everywhere, and

the force is always toward smaller values of height, i. e. towards the center of Earth.

• Sloping Down ( such as ) = force is to right or toward larger values of position. > example: for a compressed spring, potential energy is sloping down,

and the force is toward larger (less negative) values of x. • Horizontal = no force (maybe just for an instant) • If U = ME at a point where the U curve is horizontal, then there is no kinetic

energy and there is no force. This corresponds to an object at rest remaining at rest.


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February 13, 2013

SystemsIn an ideal Atwood machine two masses are released from rest. Derive an equation to solve for velocity at any point in time (using energy ONLY).

In order to solve a problem where more than one object is involved, we must do a systems analysis. We must account for the cumulative effects of each object.

Answer the following:1. What directions are the masses moving?2. Is this a conservative system?3. What is happening to their GPE?4. What is happening to their KE?5. What happens to the ME of each object?6. What is happening to the GPE, KE, and ME for the system?

Direction GPE KE MEM1

M2

System X


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February 13, 2013

SystemsIn an ideal Atwood machine two masses are released from rest. Derive an equation to solve for velocity at any point in time (using energy ONLY).

M1 = mM2 = M/V1/ = /V2/

MEsystem = ME1 + ME2

ME1 = KE1 + GPE1 + EPE1

ME2 = KE2 + GPE2 + EPE2

hA2 = 0h

hA1 = hH

H

hB2 = H

hB1 = H h

0

0

For 1 and 2VA = 0VB = V

Mass 1

MEA1 = KEA1 + GPEA1

MEA1 = mghA1 = mgh

0 Mass 1

MEA1 = KEA1 + GPEA1

MEA1 = 0

0 0


MEsystem = mgh

At point A


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February 13, 2013

Mass 1

MEB1 = KEB1 + GPEB1

MEB1 = 1/2mV2 + mg(H h)

Mass 1

MEB1 = KEB1 + GPEB1

MEB1 = 1/2MV2 MgH


MEsystem = 1/2mV2 + mg(H h) + 1/2MV2 MgH

MEsystemA = MEsystemB

mgh = 1/2mV2 + mg(H h) + 1/2MV2 MgH

mgh = 1/2mV2 + mgH mgh + 1/2MV2 MgH

0 = 1/2mV2 + mgH + 1/2MV2 MgH

MgH mgH = 1/2mV2 + 1/2MV2

(M m)gH = 1/2V2(M + m)

At point B


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February 13, 2013

(M m)gH = 1/2V2(M + m)

2(M m)gH = V2

(M + m)

2(M m)gH = V(M + m)√

Using the Big 3 and Dynamics (M m) g

VF2 = VI

2 + 2aDf

VF2 = 2aDf

VF = √2aDf

V = 2(M m)gH

(M + m)0

Df = HVf = V

(M + m)√

AMAZING(Atwood acceleration found from dynamics.)a =


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February 13, 2013

In an Atwood machine the two masses are 800 g and 400 g. The system is released from rest. How fast is the 800 g mass moving after it has fallen 120 cm

400 g

800 g


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February 13, 2013

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Work done on an object = energy gained by ... - HHS Physics