Nicholas J. Pinto and Claudio Guerra-Vela. Department of Physics and Electronics. University of Puerto Rico at Humacao. Sponsored by the National Science Foundation (NSF) All rights reserved

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Experiment 6

CONSERVATION OF ENERGY AND THE WORK-ENERGY THEOREM

In this experiment you will explore the principle of conservation of mechanical energy. You will see that gravitational energy can be converted into kinetic energy. Recall that the gravitational potential energy of an object with mass m situated at a height h from the floor (taken as the zero of energy) is given by U = mgh where g is the acceleration due to gravity. Also, the kinetic energy of an object with mass m and moving with a velocity v is given by K = mv2.

Objectives:

1. Describe the principle of the conservation of energy. 2. Apply the principle to situations involving gravitational potential energy and kinetic energy. 3. Make theoretical predictions of the final velocity of a cart after it rolls down an inclined plane

using the principle of the conservation of energy. 4. Verify the work-energy theorem 5. Compare your predictions with experimental results.

Hypothesis: How does the gravitational potential energy of a cart at the top of an inclined plane compare to the kinetic energy of the cart at the bottom of the inclined plane? How does the total energy of the cart change as it goes down the inclined plane? How does the work done on an object relate to the change in its kinetic energy?

Theory: The mechanical energy E of a system is the sum of its potential energy U and the kinetic energy K of the objects within it. In this experiment we will examine what happens to this mechanical energy when only conservative forces (like the gravitational force) cause energy transfers within the system i.e. when frictional and drag forces do not act on the objects in the system. Also, we shall assume that the system is isolated from its environment i.e. no external force from an object outside the system causes energy changes inside the system. When a conservative force does work W on an object within the system, one of these energies increases exactly as much as the other decreases. In an isolated system where only conservative forces cause energy changes, the kinetic energy and potential energy can change, but their sum, the mechanical energy E of the system cannot change. This result is called the principle of conservation of mechanical energy. We can write this principle in one more form as follows:

mechE K U = + = 0 The principle of conservation of mechanical energy allows us to solve problems that would be quite difficult to solve using only Newtons laws. When the mechanical energy of a system is conserved, we can relate the sum of the kinetic energy and potential energy at any one instant to that at another instant without considering the intermediate motion and without finding the work done by forces involved. Figure 6-1 below shows a block stationary at the top of a frictionless track at a height h from the bottom (where we assume the gravitational potential energy is zero). The mechanical energy of the block at this point is purely gravitational i.e. E = U = mgh. At the bottom of the track the block has acquired a velocity and its gravitational energy is zero. The mechanical energy of the block now is purely kinetic. i.e. E = K = mv2. At any instant on the incline, the energy of the cart would be part potential and part kinetic. The total energy of the cart however would be the same in all three positions.

Nicholas J. Pinto and Claudio Guerra-Vela. Department of Physics and Electronics. University of Puerto Rico at Humacao. Sponsored by the National Science Foundation (NSF) All rights reserved

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Figure 6-1 The block slides with no friction

Work-Energy Theorem When a body with mass m moves under the influence of a constant external force F parallel to its displacement d we say that the force has done some work. The amount of work done is given by:

W = Fd If the force applied is not along the direction of displacement of the object we need to take that component of the force along the direction of dis placement to calculate the work done the applied force. In the Figure 6-2 below, this force would be Fcos. Hence the work done in moving the block a distance d is given by:

W = Fd cos

Figure 6-2 A constant force does work while acting on a block through a distance

The general expression for the work done by a constant force F that causes a displacement s is given by:

W = F . s If the applied force is not constant during the displacement then the work done by the force needs to be calculated as follows:

f

i

x

x

W Fdx=

where F is the applied force along the x-axis and xi and xf are the initial and final positions of the mass. The work-energy theorem states that if a mass m, moves along the x axis under the influence of an external force F that points along that axis, then the work done by this force on the mass m as it moves from an initial point xi to a final point xf along the x axis is equal to the change of the kinetic energy of the mass between the two points. This can be written mathematically as:

Nicholas J. Pinto and Claudio Guerra-Vela. Department of Physics and Electronics. University of Puerto Rico at Humacao. Sponsored by the National Science Foundation (NSF) All rights reserved

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W = Kf Ki = K

If on a horizontal frictionless track a hanging mass pulls a cart that moves along the track, then the work done on the cart must equal the change in the kinetic energy of the cart. We will verify this theorem in the laboratory. See Figure 6-3

Figure 6-3 The falling weight does work on the system, which increases its kinetic energy

Examples 1. You drop a 2.0 kg textbook from a height of 10 m from the ground to a friend who stands on the

ground as shown in the Figure 6-4. (a) If the potential energy is taken as being ze ro at ground level, then what is the potential energy of the book when you release it? (b) What is the kinetic energy of the book just before your friend catches it in her outstretched hands (Hand level), which are 1.5 m above ground level. (c) How fast is the book moving as it is caught?

Book

Hand level

10 m

1.5 m

Figure 6-4 A book at rest falls from a height of 10 m. Note that the drawing is not at scale

Solution: (a) U = mgh = 2 9.8 10 = 196 J (b) Potential energy of the book at hand level is mgh = 2 9.8 1.5 = 29.4 J

Nicholas J. Pinto and Claudio Guerra-Vela. Department of Physics and Electronics. University of Puerto Rico at Humacao. Sponsored by the National Science Foundation (NSF) All rights reserved

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Since the total energy is conserved, then the kinetic energy of the book is 196 29.4 = 166.6 J (c) K = 166.6 J = mv2 .or v = 12.9 m/s

2. A frictionless roller coaster car tops the first hill in Figure 6-5 with speed v0. What is its speed at (a)

point A, (b) point B and (c) point C? (d) How high will it go on the last hill, which is too high for it to cross?

hh

h/2

voA

B

C

Figure 6-5 A frictionless roller coaster

Solution: Remember the total mechanical energy of the car is E = U + K = mgh + mv0

2 and that this energy is conserved

(a) Since point A is at the same height as that from witch the car began its fall, the speed at this point

is also vo (b) E = mg(h/2) + mv2 ,

or mgh + mvo

2 = mg(h/2) + mv2

Solving for v we get ov gh v= +2

(c) E = mv2 . Or mgh + mvo

2 = mv2

Solving for v we get ov gh v= +22

(d) At the highest point the roller coaster will come to a halt momentarily. At this height (H) the

potential energy will be the same as the kinetic energy it had at point C. i.e. Use the velocity calculated in part (c) to find the kinetic energy at point C

o

o

( )m gh v mgH

vH h

g

+ =

= +

2 2

2

12

2

2

Nicholas J. Pinto and Claudio Guerra-Vela. Department of Physics and Electronics. University of Puerto Rico at Humacao. Sponsored by the National Science Foundation (NSF) All rights reserved

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3. A 5.0 kg block moves in a straight line on a horizontal frictionless surface under the influence of a force that varies with position as shown below. See Figure 6-6. How much work does the force do as the block moves from the origin to 8.0 m?

Solution: Since the direction of the applied force is along the displacement of the block the work done

W = Fd

would just be the area under the curve shown above. We can calculate this area in four parts corresponding to the four sections A,B,C and D as seen in the Figure 6 above.

Figure 6-6 A force changes as a function of the position of an object

Part A: W = 10 2 = 20 J

Part B: Here we need the slope of this section and the form of the equation. viz. we need to know how the force varies with x (i.e. F(x) ) in this region i.e. 2 < x < 4.

F(x)= -5x + 20 (N), 2 m < x < 4 m

Hence the work done in this section is

( )W x dx J= + =4

2

5 20 10

Part C: No work is done here as F = 0.

Part D: Here we need the slope of this section and the form of the equation. viz. we need to know how the force varies with x (i.e. F(x) ) in this region i.e. 6 < x < 8.

( )F x x= +5

62

6 m < x < 8 m

Nicholas J. Pinto and Claudio Guerra-Vela. Department of Physics and Electronics. University