Download Homework #3: Conservation of Energy

Homework #3: Conservation of Energy
(I) A spring has a spring stiffness constant, k, of 440 N m . How much must this spring be stretched to store 25
J of potential energy?
2.
(I) A 7.0-kg monkey swings from one branch to another 1.2 m higher. What is the change in potential energy?
3.
(II) A 1200-kg car rolling on a horizontal surface has speed v  65 km h when it strikes a horizontal coiled
spring and is brought to rest in a distance of 2.2 m. What is the spring stiffness constant of the spring?
4.
(II) A 1.60-m tall person lifts a 2.10-kg book from the ground so it is 2.20 m above the ground. What is the
potential energy of the book relative to (a) the ground, and (b) the top of the person’s head? (c) How is the work
done by the person related to the answers in parts (a) and (b)?
5.
(II) A spring with k  53 N m hangs vertically next to a ruler. The end of the spring is next to the 15-cm mark
on the ruler. If a 2.5-kg mass is now attached to the end of the spring, where will the end of the spring line up
with the ruler marks?
6.
Find the height from which you would have to drop a ball so that it would have a speed of 9.0 m/s just before
it hits the ground.
7.
(I) Jane, looking for Tarzan, is running at top speed 5.3 m s  and grabs a vine hanging vertically from a tall
tree in the jungle. How high can she swing upward? Does the length of the vine affect your answer?
8.
(I) A novice skier, starting from rest, slides down a frictionless 35.0º incline whose vertical height is 185 m. How
fast is she going when she reaches the bottom?
9.
(I) A sled is initially given a shove up a frictionless 28.0º incline. It reaches a maximum vertical height 1.35 m
higher than where it started. What was its initial speed?
10. (II) In the high jump, Fran’s kinetic energy is transformed into gravitational potential energy without the aid of a
pole. With what minimum speed must Fran leave the ground in order to lift her center of mass 2.10 m and cross
the bar with a speed of 0.70 m s ?
1.
11.
(II) A 65-kg trampoline artist jumps vertically upward from the top of a platform with a speed of 5.0 m s . (a)
How fast is he going as he lands on the trampoline, 3.0 m below (Fig. 6–38)? (b) If the trampoline behaves like a
spring with spring stiffness constant 6.2 10 4 N m , how far does he depress it?
12.
A child and a sled with a combined mass of 50.0 kg slide down a frictionless slope. If the sled starts from rest
and has a speed of 3.00 m/s at the bottom, what is the height of the hill?
13.
A 0.250-kg block is placed on a light vertical spring (k = 5.00 × 103 N/m) and pushed downwards,
compressing the spring 0.100 m. After the block is released, it leaves the spring and continues to travel upwards. What
height above the point of release will the block reach if air resistance is negligible?
14. (II) A 62-kg bungee jumper jumps from a bridge. She is tied to a bungee cord whose unstretched length is 12 m,
and falls a total of 31 m. (a) Calculate the spring stiffness constant k of the bungee cord, assuming Hooke’s law
applies. (b) Calculate the maximum acceleration she experiences.
15.
A 70-kg diver steps off a 10-m tower and drops from rest straight down into the water. If he comes to rest 5.0
m beneath the surface, determine the average resistive force exerted on him by the water.
16.
An airplane of mass 1.5 × 104 kg is moving at 60 m/s. The pilot then revs up the engine so that the forward
thrust by the air around the propeller becomes 7.5 × 104 N. If the force exerted by air resistance on the body of the
airplane has a magnitude of 4.0 × 104 N, find the speed of the airplane after it has traveled 500 m. Assume that the
airplane is in level flight throughout this motion.
17.
A 2.1 × 103-kg car starts from rest at the top of a 5.0-m-long driveway that is inclined at 20° with the
horizontal. If an average friction force of 4.0 × 10 3 N impedes the motion, find the speed of the car at the bottom of the
driveway.
18.
A 25.0-kg child on a 2.00-m-long swing is released from rest when the ropes of the swing make an angle of
30.0° with the vertical. (a) Neglecting friction, find the child’s speed at the lowest position. (b) If the actual speed of the
child at the lowest position is 2.00 m/s, what is the mechanical energy lost due to friction?
19. (I) Two railroad cars, each of mass 7650 kg and traveling 95 km h in opposite directions, collide head-on and
come to rest. How much thermal energy is produced in this collision?
20. (II) A 21.7-kg child descends a slide 3.5 m high and reaches the bottom with a speed of 2.2 m s . How much
thermal energy due to friction was generated in this process?
21. (II) You drop a ball from a height of 2.0 m, and it bounces back to a height of 1.5 m. (a) What fraction of its
initial energy is lost during the bounce? (b) What is the ball’s speed just as it leaves the ground after the bounce?
(c) Where did the energy go?
22. (II) A 110-kg crate, starting from rest, is pulled across a floor with a constant horizontal force of 350 N. For the
first 15 m the floor is frictionless, and for the next 15 m the coefficient of friction is 0.30. What is the final speed
of the crate?
Solutions #3: Conservation of Energy
26. The elastic PE is given by PEelastic  12 kx 2 where x is the distance of stretching or
compressing of the spring from its natural length.
x
27.
2 PEelastic
k
2  25 J 

440 N m
 0.34 m
Subtract the initial gravitational PE from the final gravitational PE.

PEG  mgy2  mgy1  mg  y2  y1    7.0 kg  9.8 m s 2
29.
 1.2 m   82 J
Assume that all of the kinetic energy of the car becomes PE of the compressed spring.
2
1
2
30.
(a)
(b)
mv 2  12 kx 2

 1m s  
1200 kg   65 km h  

2
3.6 km h  
mv


 k 2 
 8.1 10 4 N m
2
 2.2 m 
x
Relative to the ground, the PE is given by

PEG  mg  ybook  yground    2.10 kg  9.80 m s 2
  2.20 m  
45.3 J
Relative to the top of the person’s head, the PE is given by

PEG  mg  ybook  yhead  h   2.10 kg  9.80 m s 2
  0.60 m   12 J
(c) The work done by the person in lifting the book from the ground to the final height is the
same
as the answer to part (a), 45.3 J. In part (a), the PE is calculated relative to the starting
location of the application of the force on the book. The work done by the person is not
related to the answer to part (b).
32. The spring will stretch enough to hold up the mass. The force exerted by the spring will be
equal to the weight of the mass.
2
mg  2.5 kg  9.80 m s
mg  k  x   x 

 0.46 m
k
53 N m


Thus the ruler reading will be 46 cm  15 cm  61cm .
5.19
With only a conservative force acting on the falling ball,
 KE  PE    KE  PE 
g
g
i
f
or
1
2
m vi2  m gyi  12 m v2f  m gyh
Applying this to the motion of the ball gives 0  m gyi  12 m v2f  0
or
yi 
v2f
2g

 9.0 m s

2
2 9.80 m s2

 4.1 m
33. The only forces acting on Jane are gravity and the vine tension. The tension
pulls in a centripetal direction, and so can do no work – the tension force is
perpendicular at all times to her motion. So Jane’s mechanical energy is
conserved. Subscript 1 represents Jane at the point where she grabs the vine, and
subscript 2 represents Jane at the highest point of her swing. The ground is the
zero location for PE  y  0  . We have v1  5.3m s , y1  0 , and v2  0 (top of
swing). Solve for y2, the height of her swing.
1
mv12  mgy1  12 mv22  mgy2  12 mv12  0  0  mgy2 
2
y2 
v12
2g

 5.3 m s 

v2 , y2
v1 , y1
2
2 9.8 m s 2

 1.4 m
No, the length of the vine does not enter into the calculation, unless the vine is less than 0.7
m long. If that were the case, she could not rise 1.4 m high. Instead she would wrap the vine
around the tree branch.
34. The forces on the skier are gravity and the normal force. The normal force is
perpendicular to the direction of motion, and so does no work. Thus the skier’s
mechanical energy is conserved. Subscript 1 represents the skier at the top of
the hill, and subscript 2 represents the skier at the bottom of the hill. The
ground is the zero location for PE  y  0  . We have v1  0 , y1  185 m , and
FN
mg 
y2  0 (bottom of the hill). Solve for v2, the speed at the bottom.
1
2
mv12  mgy1  12 mv22  mgy2  0  mgy1  12 mv22  0 


v2  2 gy1  2 9.80 m s 2 185 m   60.2 m s   135 mi h 
35. The forces on the sled are gravity and the normal force. The normal force is
perpendicular to the direction of motion, and so does no work. Thus the sled’s
mechanical energy is conserved. Subscript 1 represents the sled at the bottom
of the hill, and subscript 2 represents the sled at the top of the hill. The ground
is the zero location for PE  y  0  . We have y1  0 , v2  0 , and y2  1.35 m .
Solve for v1, the speed at the bottom.
1
mv12  mgy1  12 mv22  mgy2  12 mv12  0  0  mgy2 
2

FN
mg 

v1  2 gy2  2 9.80 m s 2 1.35 m   5.14 m s
Notice that the angle is not used in the calculation.
36. We assume that all the forces on the jumper are conservative, so that the mechanical energy
of the jumper is conserved. Subscript 1 represents the jumper at the bottom of the jump, and
subscript 2 represents the jumper at the top of the jump. Call the ground the zero location
for PE  y  0  . We have y1  0 , v2  0.70 m s , and y2  2.10 m . Solve for v1, the speed
at the bottom.
1
mv12  mgy1  12 mv22  mgy2  12 mv12  0  12 mv22  mgy2 
2
v1  v22  2 gy2 
 0.70 m s 
2

 2 9.80 m s 2
  2.10 m   6.45 m s
5.27
Since no non-conservative forces do work, we use conservation of mechanical
energy, with the zero of potential energy selected at the level of the base of the
1
1
hill. Then, m v2f  m gyf  m vi2  m gyi with yf  0 yields
2
2
yi 
v2f  vi2
2g
 3.00 m s


2
0
2 9.80 m s2

 0.459 m
Note that this result is independent of the mass of the child and sled.
5.35
Choose PEg  0 at the level of the release point and use conservation of
mechanical energy from release until the block reaches maximum height. Then,

KE f  KEi  0 and we have PEg  PEs

   PE
f
g

 PEs , or
i
1
m gym ax  0  0  kxi2 which yields
2
ym ax 


5.00  103 N m  0.100 m
kxi2

2m g
2 0.250 kg 9.80 m s2


2
 10.2 m
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.
E1  E2  12 mv12  mgy1  12 mv22  mgy2  12 mv12  mgy1  12 mv22  0 
v2   v12  2 gy1  
 5.0 m s 
2

 2 9.8 m s 2
  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  y3 . 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.
E2  E3 
1
2
1
2
mv22  mgy2  12 kx22  12 mv32  mgy3  12 kx32 
mv22  0  0  0  mgy3  12 ky32 
y3 

 mg  m 2 g 2  4  12 k   12 mv22

1
2

2  12 k 


  65 kg  9.8 m s 2 
 65 kg 
2
ky32  mgy3  12 mv22  0 

 mg  m 2 g 2  kmv22
k
9.8 m s    6.2 10
 6.2 10 N m 
2
2
4

N m  65 kg  9.154 m s 
4
 0.307 m , 0.286 m
Since y3  0 , y3  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.
E2  E3 
5.39
1
2
mv22  12 ky32  y3  v2
m
k
65 kg
  9.2 m s 
6.2 104 N m
 0.30 m
We shall take PEg  0 at the lowest level reached by the diver under the water.
The diver falls a total of 15 m, but the non-conservative force due to water
resistance acts only during the last 5.0 m of fall. The work-energy theorem then
gives

W nc  KE  PEg
or
F
av
   KE  PE 
f
g
i


cos180  5.0 m    0  0  0   70 kg 9.80 m s2 15 m  
This gives the average resistance force as Fav  2.1 103 N  2.1 kN
2
5.40
    PE 
Since the plane is in level flight, PEg
g
f
i
and the work-energy theorem
reduces to W nc  W thrust  W resistance  KE f  KEi, or
 F cos0 s  fcos180 s 12 m v
2
f
1
 m vi2
2
This gives
vf  vi2 
5.41
2 F  f s
m
2 7.5  4.0  104 N   500 m
2
60
m
s



1.5  104 kg

 77 m s
Choose PEg  0 at the level of the bottom of the driveway.

Then W nc  KE  PEg
   KE  PE 
g
f
 fcos180 s  12 m v
2
f

i
or

becomes

 0  0  m g ssin20 .

Solving for the final speed gives vf 
5.42

 2gs sin20 

vf  2 9.80 m s2  5.0 m  sin 20 

2 fs
, or
m
2 4.0  103 N
  5.0 m  
2.10  10 kg
3
3.8 m s
(a) Choose PEg  0 at the level of the bottom of the arc. The child’s initial
vertical displacement from this level is
yi   2.00 m 1 cos30.0  0.268 m
In the absence of friction, we use conservation of mechanical energy as
 KE  PE    KE  PE  , or 12 m v
g
g
f

2
f
i

 0  0  m gyi , which gives
vf  2gyi  2 9.80 m s2  0.268 m   2.29 m s
(b) With a non-conservative force present, we use

W nc  KE  PEg
   KE  PE    12 m v
g
f
i
2
f

 0   0  m gyi , or

 v2f

W nc  m   gyi 
 2



  2.00 m s2
  25.0 kg  
  9.80 m s2   0.268 m
2


  15.6 J

Thus, 15.6 J of energy is spent overcoming friction.
42. Consider this diagram for the jumper’s fall.
(a)
The mechanical energy of the jumper is conserved.
Use y
for the distance from the 0 of gravitational PE and x for the
amount of bungee cord “stretch” from its unstretched
length. Subscript 1 represents the jumper at the start of the
fall, and subscript 2 represents the jumper at the lowest
point of the fall. The bottom of the fall is the zero location
for gravitational PE  y  0  , and the location where the
bungee cord just starts to be stretched is the zero location
for elastic PE  x  0  . We have v1  0 , y1  31 m ,
Start of fall
12 m
Contact with bungee
cord, 0 for elastic PE
19 m
Bottom of fall, 0 for
gravitational PE
x1  0 , v2  0 , y2  0 , and x2  19 m . Apply conservation of energy.
E1  E2 
k
2mgy1
2
2
x

1
2
mv12  mgy1  12 kx12  12 mv22  mgy2  12 kx22  mgy1  12 kx22 

2  62 kg  9.8 m s 2
19 m 
2
  31 m   104.4 N m  1.0 10
2
N m
(b) The maximum acceleration occurs at the location of the maximum force, which
occurs when the bungee cord has its maximum stretch, at the bottom of the fall.
Write Newton’s 2nd law for the force on the jumper, with upward as positive.
Fnet  Fcord  mg  kx2  mg  ma 
a
47.
energy.
kx2
104.4 N m 19 m 
g 
 9.8 m
 62 kg 
s  22.2 m s  22 m s
m
Use conservation of energy, where all of the kinetic energy is transformed to thermal
2
2
2
2
Einitial  Efinal 
1
2
mv  Ethermal
2

 0.238 m s  
  2  7650 kg   95 km h  
 5.3  106 J


 1km h  

1
2
Fcord
mg
48. Apply the conservation of energy to the child, considering work done by gravity and work
changed into thermal energy. Subscript 1 represents the child at the top of the slide, and
subscript 2 represents the child at the bottom of the slide. The ground is the zero location for
PE  y  0  . We have v1  0 , y1  3.5 m , v2  2.2 m s , and y2  0 . Solve for the work
changed into thermal energy.
E1  E2  12 mv12  mgy1  12 mv22  mgy2  Wthermal 

Wthermal  mgy1  12 mv22   21.7 kg  9.8 m s 2
  3.5 m    21.7 kg  2.2 m s 
1
2
2
 6.9 102 J
51. (a) Calculate the energy of the ball at the two maximum heights, and subtract to find the
amount of
energy “lost”. The energy at the two heights is all gravitational PE, since the ball has
no KE at those maximum heights.
Elost  Einitial  Efinal  mgyinitial  mgyfinal
Elost

mgyinitial  mgyfinal
yinitial  yfinal

2.0 m  1.5 m
 0.25  25%
Einitial
mgyinitial
yinitial
2.0 m
(b) Due to energy conservation, the KE of the ball just as it leaves the ground is equal to its
final PE.
2
PEfinal  KEground  mgyfinal  12 mvground




vground  2gyfinal  2 9.8 m s 2 1.5 m   5.4 m s
(c) The energy “lost” was changed primarily into heat energy – the temperature of the ball
and the ground would have increased slightly after the bounce. Some of the energy
may have been changed into acoustic energy (sound waves). Some may have been lost
due to non-elastic deformation of the ball or ground.
52. Since the crate moves along the floor, there is no change in gravitational
PE, so use the work-energy theorem: Wnet  KE2  KE1 . There are two forces
Ffr
FP
doing work: FP , the pulling force, and Ffr  k FN  k mg , the frictional
force. KE1  0 since the crate starts from rest. Note that the two forces doing
work do work over different distances.
WP  FP d P cos 0o
Wfr  Ffr d fr cos180o    k mgd fr
Wnet  WP  Wfr  KE2  KE1  12 mv22  0 
v2 

2
m
WP  Wfr  
2
m
 FP d P   k mgd fr 
 350 N  30 m    0.30 110 kg   9.8 m s 2  15 m    10 m s
110 kg 
2
FN
mg