1. No category

Ch.8 #63,65,77,78,81,87 Solutions 63. Picture the Problem: You and your friend each solve the same physics problem involving the motion of a skier down a slope, but choose different locations to correspond to y = 0. Strategy: Note that only changes in the gravitational potential energy mgy are important for predicting speeds in physics
problems like this one, not the absolute value of the gravitational potential energy.
Solution: 1. (a) The choice of the location of y = 0 affects the absolute value of the potential energy so that your answer and your friend’s answer will disagree on this quantity. 2. (b) The choice of the location of y = 0 does not affect the calculated change in potential energy so that your answer and your friend’s answer will agree on this quantity. 3. (c) The choice of the location of y = 0 does not affect the calculated change in potential energy. Since the kinetic energy depends only on change of potential energy, your answer and your friend’s answer will agree on this quantity. Insight: From the standpoint of predicting the skier’s speed, it makes no difference whether the skier’s potential energy changes from 1075 J to 1000 J or from 575 J to 500 J. The skier gains 75 J of kinetic energy in either case. 65. Picture the Problem: A leaf falls to the ground with constant speed.
Strategy: Consider the value of the mechanical energy of the leaf in order to answer the question.
Solution: Since the leaf’s speed does not change its kinetic energy remains constant. Its gravitational potential energy, however, decreases as the leaf descends to a lower elevation. We conclude that the value of Ki + Ui is greater than the value of Kf + Uf. Insight: As the leaf falls air friction is changing its mechanical energy into thermal energy (it heats up the air slightly).
77. Picture the Problem: The child slides from rest at point A and lands at point B as indicated in the figure at right. Strategy: Use the conservation of mechanical energy to find the horizontal speed of the child at the bottom of the slide in terms of h. Then use equation 4‐9, the landing site of a projectile launched horizontally, to find the speed the child should have in order to land 2.50 m down range. Set the speeds equal to each other and solve for h. K A  U A  K bottom  U bottom
Solution: 1. Use equation 8‐3 and let EA  Ebottom to find vbottom : 1
2
2
 mgybottom
mv  mgyA  12 mvbottom
2
A
2
 mg 1.50 m 
0  mg  h  1.50 m   12 mvbottom
2 gh  vbottom
x  vbottom 2 ybottom g  vbottom  x g 2 ybottom 2. Use equation 4‐9 to find the appropriate vbottom for the child to land 2.50 m down range: 2 gh  x g 2 ybottom
3. Set the two velocities equal to each other and solve for h: 2 gh  x 2 g 2 ybottom
 2.50 m 
x2

 1.04 m
4 ybottom 4 1.50 m 
2
h
Insight: These are the sort of calculations an engineer might make to determine how high to build a slide so that a child will land in a certain place. However, an engineer should account for the nonzero friction when making his design. 78. Picture the Problem: The child slides from rest at point A and lands at point B as indicated in the figure at right. Strategy: Use the conservation of mechanical energy to find the horizontal speed of the child at the bottom of the slide. Then use equation 4‐9 to find the landing site of the child because she is launched horizontally. K A  U A  K bottom  U bottom
Solution: 1. Use equation 8‐3 and let EA  Ebottom to find vbottom : 1
2
1
2
2
 mgybottom
mv  mgyA  12 mvbottom
2
A
2
 mg 1.50 m 
mvA2  mg  h  1.50 m   12 mvbottom
vA2  2 gh  vbottom
2. Use equation 4‐9 to find x: x  vbottom 2 ybottom g 
v
2
A
 2 gh   2 ybottom g 
2
  0.54 m/s   2  9.81 m/s 2   3.2 m    2 1.50 m   9.81 m/s 2   

x  4.4 m
Insight: Her initial speed doesn’t make much difference in the landing spot; verify for yourself that she lands at 4.39 m with the initial speed as opposed to 4.38 m when starting from rest. This is quite a slide—4.4 m is 14 ft from the base of the slide, and the intrepid swimmers hit the water at 9.6 m/s (21 mi/h)! 81. Picture the Problem: The physical situation is depicted in the figure.
Strategy: Use the conservation of mechanical energy and the geometry of the problem to find the speed of the skateboarder at point A. Let yA  0 and vB  0 .
Solution: Set EA  EB to find vA : KA  UA  KB  U B
mvA2  0  0  mgyB
1
2
vA  2 gyB  2  9.81 m/s 2   2.64 m 
 7.20 m/s
Insight: The solution assumes there is no friction as the skateboarder travels around the half pipe. In real life the skater’s speed must exceed 7.20 m/s at point A because friction will convert some of his kinetic energy into heat. 87. Picture the Problem: The block slides from rest at point A and is launched horizontally at point B as indicated in the figure at right. Strategy: Use equation 8‐9, Wnc   K B  U B    K A  U A  , to find the change in mechanical energy due to the presence of friction. Use the resulting expression to find the horizontal speed of the block at the bottom of the ramp. Then use equation 4‐9 to find the landing site of the block because it is launched horizontally. Solution: 1. Write equation 8‐9 and solve for vB : K A  U A  Wnc  K B  U B
0  mgyA  Wnc  12 mvB2  mgyB 2 g  yA  yB   2 Wnc m  vB
2. Use equation 4‐9 to find d: d  vB 2 yB g   2 g  yA  yB   2 Wnc m   2 yB g   4 yB  yA  yB   Wnc mg 


 9.7 J 

 4  0.25 m  1.50  0.25 m  
2

1.9 kg   9.81 m/s  
d  0.85 m
Insight: The block does not travel as far because friction has converted some of the kinetic energy into heat.