CHAPTER 6: Work and Energy
Answers to Questions
2. Since “centripetal” means “pointing to the center of curvature”, then a centripetal force will not do work on an object, because if an object is moving in a curved path, by definition the direction towards the center of curvature is always perpendicular to the direction of motion. For a force to do work, the force must have a component in the direction of displacement. So the centripetal force does no work.
5. The kinetic force of friction opposes the relative motion between two objects. As in the example suggested, as the tablecloth is pulled from under the dishes, the relative motion is for the dishes to be left behind as the tablecloth is pulled, and so the kinetic friction opposes that and moves the dishes in the same direction as the tablecloth. This is a force that is in the direction of displacement, and so positive work is done. Also note that the cloth is moving faster than the dishes in this case, so that the friction is kinetic, not static.
7. (a) In this case, the same force is applied to both springs. Spring 1 will stretch less, and so more work is done on spring 2. (b) In this case, both springs are stretched the same distance. It takes more force to stretch spring 1, and so more work is done on spring 1.
Solutions to Problems
3. (a) See the free-body diagram for the crate as it is being pulled. Since the Dx crate is not accelerating horizontally, F= F = 230 N . The work done to P fr F F move it across the floor is the work done by the pulling force. The angle fr P o between the pulling force and the direction of motion is 0 . o 2 mg FN WP= F P d cos0 =( 230 N)( 4.0 m)( 1) = 9.2 10 J
(b) See the free-body diagram for the crate as it is being lifted. Since the crate is not Dy accelerating vertically, the pulling force is the same magnitude as the weight. The FP angle between the pulling force and the direction of motion is 0o. W= F dcos 0o = mgd = 1300 N 4.0 m = 5.2 10 3 J P P ( )( ) mg
F 8. The piano is moving with a constant velocity down the plane. FP is the N F force of the man pushing on the piano. fr nd y (a) Write Newton’s 2 law on each direction for the piano, with an FP acceleration of 0. x mg Fy = FN - mgcosq = 0 � F N mg cos q
Fx = mgsinq - FP - F fr = 0
FP= mgsinq - F fr = mg ( sin q - mk cos q ) =( 330 kg)( 9.80 m s2)( sin 28 o - 0.40cos 28 o) = 3.8 10 2 N (b) The work done by the man is the work done by FP . The angle between FP and the direction of motion is 180o. W= F d cos180o = -( 380 N)( 3.6 m) = - 1.4 10 3 J . P P o (c) The angle between Ffr and the direction of motion is 180 .
o 2 o Wfr= F fr dcos180 = -mk mgd cos q = -( 0.40)( 330 kg)( 9.8m s)( 3.6 m) cos 28 . = - 4.1 103 J (d) The angle between the force of gravity and the direction of motion is 62o. So the work done by gravity is
o o 2 o 3 WG= F G dcos 62 = mgd cos 62 =( 330 kg)( 9.8 m s)( 3.6 m) cos 62 = 5.5 10 J . (e) Since the piano is unaccelerated, the net force on the piano is 0, and so the net work done on the piano is also 0. This can also be seen by adding the three work amounts calculated.
WNet= W P + W fr + W G = -1400 J - 4100 J + 5500 J = 0 J
12. The work done will be the area under the Fx vs. x graph. (a) From x = 0.0 to x = 10.0 m , the shape under the graph is trapezoidal. The area is 10 m+ 4 m W =(400 N) = 2.8 103 J a 2 (b) From x = 10.0 m to x = 15.0 m , the force is in the opposite direction from the direction of motion, and so the work will be negative. Again, since the shape is trapezoidal, we find 5 m+ 2 m W =( -200 N) = - 700 J . a 2 Thus the total work from x = 0.0 to x = 15.0 m is 2800 J- 700 J = 2.1 103 J
1 2 16. (a) Since KE= 2 mv , then v= 2( KE) m and so v KE . Thus if the kinetic energy is doubled, the speed will be multiplied by a factor of 2 .
1 2 2 (b) Since KE= 2 mv , then KE v . Thus if the speed is doubled, the kinetic energy will be multiplied by a factor of 4 . 37. (a) Since there are no dissipative forces present, the mechanical energy of the person – trampoline – Earth combination will be conserved. The level of the unstretched trampoline is the zero level for both the elastic and gravitational PE. Call up the positive direction. Subscript 1 represents the jumper at the top of the jump, and subscript 2 represents the jumper upon arriving at the trampoline. There is no elastic PE involved in this part of
the problem. We have v1 = 5.0 m s , y1 = 3.0 m , and y2 = 0 . Solve for v2, the speed upon arriving at the trampoline. 12 1 2 1 2 1 2 E1= E 2 � 2 mv 1 = mgy 1 + 2 mv 2 � mgy 2 = 2 mv 1 + mgy 1 2 mv 2 0
22 2 v2=� v 1 =2 gy 1 � =( 5.0 m s) 被 2( 9.8 m s)( 3.0 m) 9.154 m s 9.2 m s
The speed is the absolute value of v2 . (b) Now let subscript 3 represent the jumper at the maximum stretch of the trampoline. We
have v2 = 9.154 m s , y2 = 0 , x2 = 0 , v3 = 0 , and x3= y 3 . There is no elastic energy at position 2, but there is elastic energy at position 3. Also, the gravitational PE at
position 3 is negative, and so y3 < 0 . A quadratic relationship results from the conservation of energy condition. 12 1 2 1 2 1 2 E2= E 3 � 2 mv 2 + mgy 2 = 2 kx 2 + 2 mv 3 + mgy 3 2 kx 3
12 1 2 1 2 1 2 2mv2+0 + 0 = 0 + mgy 3 + 2 ky 3 � 2 ky 3 - mgy 3 = 2 mv 2 0
2 21 1 2 2 2 2 -mg� m g - 4( 2 k)( 2 mv2 ) -mg� m g kmv y = = 2 3 1 2( 2 k) k
22 2 -(65 kg)( 9.8 m s2) � ( 65 kg) ( 9.8m s 2) ( 6.2 10 4 N m)( 65 kg)( 9.154 m s) = (6.2 104 N m) = - 0.307 m , 0.286 m
Since y3<0 , y 3 = - 0.31 m . The first term under the quadratic is about 1000 times smaller than the second term, indicating that the problem could have been approximated by not even including gravitational PE for the final position. If that approximation would have been made, the result would have been found by taking the negative result from the following solution. m 65 kg E= E � 1 mv2 � 1 ky 2 = y = v ( 9.2 m s) 0.30 m 2 32 2 2 3 3 2 k 6.2 104 N m
45. The maximum acceleration of 5.0 g occurs where the force is at a maximum. The maximum force occurs at the bottom of the motion, where the spring is at its maximum compression. Write Newton’s 2nd law for the elevator at the bottom of the mg F motion, with up as the positive direction. spring
Fnet= F spring - Mg = Ma = 5.0 Mg � F spring 6.0 Mg Now consider the diagram for the elevator at various points in its motion. If there are no non-conservative forces, then mechanical energy is conserved. Subscript 1 represents the Start of fall elevator at the start of its fall, and subscript 2 represents the h elevator at the bottom of its fall. The bottom of the fall is Contact with the zero location for gravitational PE ( y = 0) . There is spring, x also a point at the top of the spring that we will define as 0 for elastic PE Bottom of fall, 0 for the zero location for elastic PE (x = 0). We have v = 0 , 1 gravitational PE
y1 = x + h , x1 =0 , v2 = 0 , y2 = 0 , and x2 = x . Apply conservation of energy.
12 1 2 1 2 1 2 E1= E 2 � 2 Mv 1 + Mgy 1 = 2 kx 1 + 2 Mv 2 + Mgy 2 2 kx 2
12 1 2 0+Mg( x + h) + 0 = 0 + 0 +2 kx2 � Mg( x = h) 2 kx 2 6.0Mg骣 6 Mg 骣 6 Mg2 12 Mg F=6.0 Mg = kx � x � = Mg琪 � h1 k 琪 k spring k桫 k2 桫 k h
52. Since the crate moves along the floor, there is no change in gravitational Ffr F PE, so use the work-energy theorem: Wnet= KE 2 - KE 1 . There are two forces P doing work: F , the pulling force, and F=m F = m mg , the frictional P frk N k F mg force. KE1 = 0 since the crate starts from rest. Note that the two forces doing N work do work over different distances. o o WP= F P d Pcos0 W fr = F fr d fr cos180 = -mk mgd fr 1 2 Wnet= W P + W fr = KE 2 - KE 1 =2 mv 2 - 0 2 2 v=( W + W) =( F d - m mgd ) 2m P fr m P Pk fr 2 =轾( 350 N)( 30 m) -( 0.30)( 110 kg)( 9.8m s2 )( 15 m) = 10 m s (110 kg) 臌
58. The work necessary to lift the piano is the work done by an upward force, equal in magnitude to the weight of the piano. Thus W= Fdcos 0o = mgh . The average power output required to lift the piano is the work done divided by the time to lift the piano. 2 W mgh mgh (315 kg)( 9.80 m s)( 16.0 m) P= = � t = = 28.2 s t t P 1750 W
J6: What is the speed of a 2.0 kg mass, such that 4.0 Joules of work are required to double it?