HW 04 - fa14 - work and energy: Chapter 7: 1, 11, 16, 25, 30, 42, 67. Chapter 8: 6, 8, 23, 24, 36, 38, 62.

Pool of problems to study for quiz: 7.1, 7.25, 7.42, 8.6, 8.8, 8.23, 8.36, 8.62.

Chapter 7, problem 1 (9th ed) Chapter 7, problem 1 (10th ed): A proton (mass ) is being accelerated along a straight line at in a machine. If the proton has an initial speed of and travels , what then is (a) its speed and (b) the increase in its kinetic energy?

Let the straight line from the problem statement be in the +x-direction. We can verify that kinematics and the work energy theorem in which , , and yield the same final speed,

Hence, the exact symbolic result matches the result obtain by simple use of . Plugging in numbers, one should get . The increase in kinetic energy is given by the work,

Note that is a new, frequently-encountered SI unit called the Joule (after James Prescott Joule).

Chapter 7, problem 11 (9th ed) Chapter 7, problem 11 (10th ed): A force with a fixed orientation (i.e., ) does work on a particle as the particle moves through displacement. What is the angle between the force and the displacement if the change in the particle's kinetic energy is (a) and (b) ?

Chapter 7, problem 16 (9th ed) Chapter 7, problem 16 (10th ed): An object is moving in the positive direction of an x-axis. When it passes through, a constant force directed along the axis begins to act on it. Figure 7-28 gives its kinetic energy K versus position x as it moves from to ;. The force continues to act. What is v when the object moves back through?

Solution: The change in kinetic energy is computed to eventually compute . Note that the change in kinetic energy is negative; i.e., the object is slowing down.

We expect, therefore, a negative acceleration (since the object was moving in the positive x-direction). Using the kinematic statement (again, derivable from the work-energy theorem),

However, we need an initial velocity from which the object accelerates. When , the kinetic energy becomes zero, implying that the mass comes to rest momentarily. Thus,

The speed of the object when is,

Chapter 7, Problem 25 (9th ed) Chapter 7, Problem 25 (10th ed): In the figure, a m = 0.250 kg block of cheese lies on the floor of a M = 900 kg elevator cab that is being pulled (accelerated!) upward by a cable through distance d = 2.40 m. If the normal force on the block from the floor has constant magnitude FN = 3.00 N, how much work is done on the cab by the force from the cable?

The acceleration of the elevator is unknown, but can be found out from the mass of the cheese, and the normal force the floor of the elevator is exerting upon it. The net upward force is given by applying Newtons 2nd Law applied to both the cheese and the elevator-car, in which the acceleration of the cheese equals the acceleration of the elevator (conservation of string again). Let be the force that accelerates the cheese-elevator system upwards. Then,

In , the force-sum is given by,

We also apply Newtons 2nd Law to the cheese alone: the two forces acting upon the cheese are and , which is such that the cheese has a positive acceleration of ,

Thus the force from the cable is given by equating and ,

The work done by the cable on the cab, subsequently, is,

N.B.: I just want to point out that the picture of the elevator and the block of cheese is NOT a schematic. Why not? Because no forces are labeled, neither is a coordinate system indicated, or units, etc. Thats what distinguishes a picture from a schematic. The above illustration of a yellow block in an elevator is about as much a schematic as various pictures of Grumpy Cat are.

** Chapter 7, problem 30 (9th ed) ** Chapter 7, problem 30 (10th ed): A block of mass m lies on a horizontal frictionless surface and is attached to one end of a horizontal spring (spring constant k) whose other end is fixed. The block is initially at rest at the position where the spring is unstretched (x=0) when a constant horizontal force in the positive direction of the x axis is applied to it. A plot of the resulting kinetic energy of the block versus its position x is shown in Fig. 7-35. The scale of the figures vertical axis is set by Ks=4.0 J. (a) What is the magnitude of F? (b) What is the value of k?

You have two simultaneous equations from this plot, of the general expression , which we solve by using the general formula for the 2x2 inverse, , and

Chapter 7, Problem 42 (9th ed) Chapter 7, Problem 42 (10th ed): The figure shows a cord attached to a cart that can slide along a frictionless horizontal rail aligned along an x axis. The left end of the cord is pulled over a pulley, of negligible mass and friction and at cord height h = 1.20 m, so the cart slides from x1 = 3.00 m to x2 = 1.00 m. During the move, the tension in the cord is a constant 25.0 N. What is the change in the kinetic energy of the cart during the move?

The work-energy theorem is used, but one must use the identity upon (in which I illustrated the angle in the schematic), and eliminate the trig-functions in favour of the dimensions illustrated above. Then, one uses the antiderivative , where is the constant of integration, and write,

Chapter 7, Problem 62 (9th ed) Chapter 7, Problem 62 (10th ed) A spring with a pointer attached is hanging next to a scale marked in millimeters. Three different packages are hung from the spring, in turn, as shown,

Introduce the notation,

(a) Which mark on the scale will the pointer indicate when no package is hung from the spring? we have two invocations of hookes law for the and packages. Using the notation ,

Dividing the two equations eliminates the unknown , and provides an expression that can be solved for . Subtracting the two equations eliminates the unknown , and provides an expression that can be solved for which, in turn, is needed to calculate (and thats essentially the whole problem),

(b) What is the weight of the third package?

As mentioned before, we need the spring-constant,

Using Newtons 2nd Law in exactly the same manner as , except this time with the known quantities and , we have,

Introducing the symbolic expressions for to obtain an exact answer in problem givens (which we enumerated in ),

Chapter 8, problem 6 (9th ed) Chapter 8, problem 6 (10th ed): In Fig. 8-31, a small block of mass m = 0.032 kg can slide along the frictionless loop-the-loop, with loop radius R = 12 cm. The block is released from rest at point P, at height h = 5.0R above the bottom of the loop. How much work does the gravitational force do on the block as the block travels from point P to (a) point Q and (b) the top of the loop? If the gravitational potential energy of the block Earth system is taken to be zero at the bottom of the loop, what is that potential energy when the block is (c) at point P, (d) at point Q, and (e) at the top of the loop? (f ) If, instead of merely being released, the block is given some initial speed down-ward along the track, do the answers to (a) through (e) increase, decrease, or remain the same?

Solution: (a) The displacement between the initial point and Q has a vertical component of h R downward (same direction as ), so (with h = 5R) we obtain,

(b) The displacement between the initial point and the top of the loop has a vertical component of h 2R downward (same direction as ), so (with h = 5R) we obtain,

(c) With y = h = 5R, at P we find

.

(d) With y = R, at Q we have

.

(e) With y = 2R, at the top of the loop, we find

.

(f) The new information is not involved in any of the preceding computations; the above results are unchanged. Note that this means work is a quantity insensitive to a particles initial and final state; it is something that happens along the way.

NOTE: I would like to present a varied solution to the above problem.

sliding mass in loop and reaction force: A small block of mass m starts from rest and slides along a frictionless loop-the-loop as shown in the left-hand figure on the top of the next page,

What should be the initial height z, so that m pushes against the top of the track (at a) with a force equal to its weight?

We sum the forces at the top of the loop-the-loop, which subsequently tells us the velocity at the top of the loop-the-loop,

The work-energy theorem, therefore, says that gravity must do an amount of work such that the block has this velocity ,

Solving the equation for using the quantities in , we have,

chapter 8, problem 8 (9th ed) chapter 8, problem 8 (10th ed): A snowball is fired from a cliff high. The snowballs initial velocity is, directed above the horizontal. (a) How much work is done on the snowball by the gravitational force during its flight to the flat ground below the cliff? (b) What is the change in the gravitational potential energy of the snowballEarth system during the flight? (c) If that gravitational potential energy is taken to be zero at the height of the cliff, what is its value when the snowball reaches the ground? Solution: Let the y-coordinate of the cliff be zero: . Let the snowball reach a maximum height of , then fall a displacement of , and lastly fall a displacement of (after which it splats at the bottom of the cliff).

(a) The work done by gravity is divided up into that done over these three legs of the trip,

(b) Defining the system as the Earth/snowball, we see that potential energy is lost, and so , where is the kinetic energy gained (by the Earth-snowball (specifically, snowball)) by the indicated loss of gravitational potential energy.

(c) This is a bit of a silly question, but it helps us appreciate that potential energy is ambiguous to within a constant[footnoteRef:1], . The potential energy when it reaches the ground is less than the potential energy when it is fired by |U|, so U = 184 J when the snowball hits the ground. [1: As opposed to being ambiguous to within a linear function of , , quadratic function of , , exponential function of , , etc., where is a spatial coordinate (potential energy gains meaning only insofar as it varies in space and/or time).]

chapter 8, problem 23 (9th ed) chapter 8, problem 23 (10th ed): The string in Fig. 8-36 is L = 120 cm long, has a ball attached to one end, and is fixed at its other end. The distance d from the fixed end to a fixed peg at point P is d = 75.0 cm. When the initially-stationary ball is released with the string horizontal as shown, it will swing along the dashed arc. What is its speed when it reaches (a) its lowest point and (b) its highest point after the string catches on the peg?

Solution: (a) As the string reaches its lowest point, its original potential energy U = mgL (measured relative to the lowest point) is converted into kinetic energy. Thus,

The mass of the ball is not needed, so what we were really working with was kinetic and potential energy per unit mass.

(b) In this case, the total mechanical energy is shared between kinetic and potential mgyb. We note that yb = 2r where r = L d = 0.450 m. Energy conservation leads to,

chapter 8, problem 24 (9th ed.) chapter 8, problem 24 (10th ed.): A block of mass m = 2.0 kg is dropped from height h = 40 cm onto a spring of spring constant k = 1960 N/m (picture on left). Find the maximum distance the spring is compressed.

Let down be positive, and let be the fully-compressed spring-length, implying that the block is dropped from . The block is of mass , dropped from above the springs relaxed length, and the spring is subsequently compressed a (positive) displacement . Consequently, the block drops a total (positive) distance . Conservation of energy, , yields,

We note that the quantities are dimensionless; we demonstrate this,

This helps clean up the notation a bit. Using the quadratic formula, implies,

Just to show you that you get the same answer,

chapter 8, problem 36 (9th ed) chapter 8, problem 36 (10th ed): Two children are playing a game in which they try to hit a small box on the floor with a marble fired from a spring-loaded gun that is mounted on a table. The target box is horizontal distance D = 2.20 m from the edge of the table; see Fig. 8-46. Bobby compresses the spring, but the center of the marble falls short of the center of the box. How far should Rhoda compress the spring to score a direct hit? Assume that neither the spring nor the ball encounters friction in the gun.

Solution: If the spring constant is , and the height of the table is (neither are givenwe are introducing it as a placeholder-variable!), then Rhoda must compress the spring a distance , given by the following (in which we introduce and time-of-flight as placeholder variables),

Using the same machinations as , Bobby compressed the spring a distance given by,

In both and , we have the (matching!) time-of-flight of both marbles,

So we have many unknown placeholder variables. As I mentioned in class, taking the ratio of two expressions bearing the same units can often produce en-masse (and quite satisfying) cancellations,

chapter 8, problem 38 (9th ed.) chapter 8, problem 38 (10th ed.): Consider a plot of potential energy U versus position x of a particle that can travel only along an x-axis under the influence of a conservative force. The graph has these values: , and . The particle is released at the point where U forms a potential hill of height , with kinetic energy. What is the speed of the particle at (a) and (b) ? What is the position of the turning point on (c) the right side and (d) the left side?

Solution: In this problem, the mechanical energy (the sum of K and U) remains constant as the particle moves.

(a) using conservation of energy, in which and , in which , , and , we can calculate the velocity at point-A, , due to ,

(b) repeating the same procedure as part-a, except using , in which and ,

(c) Turning points are where the total energy equals potential energy (equivalent to having zero velocity). We calculate the total energy of the system is . The potential energies with such a height must be determined by interpolation. Graphically, this interpolation appears as,

Algebraically, this interpolation is,

Chapter 8, Problem 62 (9th ed) Chapter 8, Problem 62 (10th ed): In Fig. 8-53, a block slides along a path that is without friction until the block reaches the section of length, which begins at height on a ramp of angle . In that section, the coefficient of kinetic friction is. The block passes through point A with a speed of. If the block can reach point B (where the friction ends), what is its speed there, and if it cannot, what is its greatest height above A?

The particles mass is not given; however, we can divide both sides of energy conservation by the objects mass. Let be the energy just before entering the friction-laden region, be the energy at point-A, and the energy at point-B. Dividing energy conservation by mass,

Let be the work done by friction in the region of length (below) point-B, on which the block experiences a force in the normal direction (in inclined-plane coordinates); to wit, the normal force. Then,

The energy per unit mass in each region is then given by the following, in which gravitational energy is spent in ascending to the onset of friction (), and up the friction-laden trip (),

If gravitational potential energy is defined to be 0 at point-A, then a negative energy per unit mass is impossible, and if our numerical computations of produce a negative number, we see we have a forbidden region,

Hence, the block does indeed make it up the plane. The speed at B, then, is,