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DATE _______________ CLASS ____________________
Holt Physics
Problem 5E
CONSERVATION OF MECHANICAL ENERGY
PROBLEM
The largest apple ever grown had a mass of about 1.47 kg. Suppose you
hold such an apple in your hand. You accidentally drop the apple, then
manage to catch it just before it hits the ground. If the speed of the apple
at that moment is 5.42 m/s, what is the kinetic energy of the apple? From
what height did you drop it?
SOLUTION
1. DEFINE
2. PLAN
Given:
m = 1.47 kg
g = 9.81 m/s2
v = 5.42 m/s
Unknown:
KE = ?
h=?
Choose the equation(s) or situation: Use the equations for kinetic and
gravitational potential energy.
1
KE =  mv 2
2
PEg = mgh
The zero level for gravitational potential energy is the ground. Because the apple
ends its fall at the zero level, the final gravitational potential energy is zero.
PEg,f = 0
PEg,i = mgh
The apple is initially at rest, so the initial kinetic energy is zero.
KEi = 0
The final kinetic energy is therefore:
1
KEf =  mv 2
2
3. CALCULATE
Substitute the values into the equation(s) and solve:
2)h
PEg,i = (1.47 kg)(9.81 m
s
1
m2
KEf = (1.47 kg) 5.42 
2
s
Solving for KE yields the following result:
1
m2
KE = KEf = (1.47 kg) 5.42  = 21.6 J
2
s
50
Holt Physics Problem Workbook
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The initial gravitational potential energy is measured at the point from which the
apple is released.
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Now use the principle of conservation for mechanical energy and the calculated
quantity for KEf to evaluate h.
MEi = MEf
PEi + KEi = PEf + KEf
PEi + 0 J = 0 J + 21.6 J
mgh = 21.6 J
21.6 J
h =  = 1.50 m
2
(1.47 kg) 9.81 m
s
4. EVALUATE
Note that the height of the apple can be determined without knowing the apple’s
mass. This is because the conservation equation reduces to the equation relating
speed, acceleration, and displacement: v 2 = 2gh.
ADDITIONAL PRACTICE
Copyright © by Holt, Rinehart and Winston. All rights reserved.
1. The largest watermelon ever grown had a mass of 118 kg. Suppose this
watermelon were exhibited on a platform 5.00 m above the ground. After
the exhibition, the watermelon is allowed to slide along to the ground
along a smooth ramp. How high above the ground is the watermelon at
the moment its kinetic energy is 4.61 kJ?
2. One species of eucalyptus tree, Eucalyptus regnens, grows to heights similar to those attained by California redwoods. Suppose a bird sitting on
top of one specimen of eucalyptus tree drops a nut. If the speed of the
falling nut at the moment it is 50.0 m above the ground is 42.7 m/s, how
tall is the tree? Do you need to know the mass of the nut to solve this
problem? Disregard air resistance.
3. In 1989, Michel Menin of France walked on a tightrope suspended under
a balloon nearly at an altitude of 3150 m above the ground. Suppose a
coin falls from Menin’s pocket during his walk. How high above the
ground is the coin when its speed is 60.0 m/s?
4. In 1936, Col. Harry Froboess of Switzerland jumped into the ocean from
the airship Graf Hindenburg, which was 1.20 × 102 m above the water’s
surface. Assuming Froboess had a mass of 72.0 kg, what was his kinetic
energy at the moment he was 30.0 m from the water’s surface? What was
his speed at that moment? Neglect the air resistance.
5. Suppose a motorcyclist rides a certain high-speed motorcycle. He reaches
top speed and then coasts up a hill. The maximum height reached by the
motorcyclist is 250.0 m. If 2.55 × 105 J of kinetic energy is dissipated by
friction, what was the initial speed of the motorcycle? The combined
mass of the motorcycle and motorcyclist is 250.0 kg.
6. The deepest mine ever drilled has a depth of 12.3 km (by contrast,
Mount Everest has height of 8.8 km). Suppose you drop a rock with a
Problem 5E
51
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mass of 120.0 g down the shaft of this mine. What would the rock’s kinetic energy be after falling 3.2 km? What would the potential energy associated with the rock be at that same moment? Assume no air resistance
and a constant free-fall acceleration.
Copyright © by Holt, Rinehart and Winston. All rights reserved.
7. Desperado, a roller coaster built in Nevada, has a vertical drop of 68.6 m.
The roller coaster is designed so that the speed of the cars at the end of
this drop is 35.6 m/s. Assume the cars are at rest at the start of the drop.
What percent of the initial mechanical energy is dissipated by friction?
52
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Givens
Solutions
10. ∆x = 9.50 × 102 m
∆x
a. vx = vi (cos q) = 
∆t
∆x
∆t = 
vi(cos q)
q = 45.0°
m = 65.0 g
g = 9.81 m/s2
∆t
vertical speed of the arrow for the first half of the flight = vi (sin q) = g 
2
g∆x


vi (sin q) =
2vi(cos q)
(9.81 m/s )(9.50 × 10 m)
 = 96.5 m/s

2(singq∆)
(xcos q ) = (2)(sin 45.0°)(cos 45.0°)
2
vi =
2
KEi = 2 mvi2 = 2 (65.0 × 10−3 kg)(96.5 m/s)2 = 303 J
1
x = 55.0 cm
1
b. From the conservation of energy,
PEelastic = KEi
1
 kx 2
2
= KEi
2KEi
(2)(303 J)
k = 
= 
= 2.00 × 103 N/m
2
x
(55.0 × 10−2 m)2
II
c. KE i = PEg,max + KE f
KE f = 2 mvx 2 = 2 m[(vi (cos q)]2 = 2 (65.0 × 10−3 kg)(96.5 m/s)2(cos 45.0°)2 = 151 J
1
1
1
PEg,max = KE i − KE f = 303 J − 151 J = 152 J
PEg ,max
152 J
hmax = 
=  
mg
(65.0 × 10−3 kg)(9.81 m/s2)
h = 238 m
Copyright © by Holt, Rinehart and Winston. All rights reserved.
Additional Practice 5E
1. m = 118 kg
PEi + KEi = PEf + KEf
1
hi = 5.00 m
mghi + 2 mvi2 = mghf + KEf
g = 9.81 m/s2
mghf = mghi + 2 mvi2 − KEf
vi = 0 m/s
KEf = 4.61 kJ
1
(0 m/s)2
4.61 × 103 J
v 2 KEf
hf = hi + i −  = 5.00 m + 
− 
2
(2)(9.81 m/s )
(118 kg)(9.81 m/s2)
2g mg
hf = 5.00 m − 3.98 m = 1.02 m above the ground
2. vf = 42.7 m/s
PEi + KEi = PEf + KEf
1
hf = 50.0 m
vi = 0 m
g = 9.81 m/s2
1
mghi + 2mvi 2 = mghf + 2mvf 2
(42.7 m/s)2 − (0 m/s)2
vf2 − vi2
hi = hf + 
= 50.0 m + 
= 50.0 m + 92.9 m
(2)(9.81 m/s2)
2g
hi = 143 m
The mass of the nut is not needed for the calculation.
Section Two — Problem Workbook Solutions
II Ch. 5–7
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Givens
Solutions
PEi + KEi = PEf + KEf
3. hi = 3150 m
1
mghi = mghf + 2mvf 2
(60.0 m/s)2
vf 2
hf = hi −  = 3150 m − 
= 3150 m − 183 m
(2)(9.81 m/s2)
2g
vf = 60.0 m/s
KEi = 0 J
g = 9.81 m/s2
hf = 2970 m
PEi + KEi = PEf + KEf
4. hi = 1.20 × 102 m
PEi − PEf = KEf
hf = 30.0 m
KEf = ∆PE = mg(hi − hf )
m = 72.0 kg
g = 9.81 m/s
KEf = (72.0 kg)(9.81 m/s2)(1.20 × 102 m − 30.0 m) = (72.0 kg)(9.81 m/s2)(9.0 × 101 m)
KEi = 0 J
KEf = 6.4 × 104 J
2
vf =
2KEf
 =
m
(2)(6.4 × 104 J)

72.0 kg
vf = 42 m/s
II
5. hf = 250.0 m
1
5
∆ME = −2.55 × 10 J
m = 250.0 kg
2
g = 9.81 m/s
∆ME = PEf − KEi = mghf − 2mvi2
vi =
2∆ ME
2ghf −   =
m
(2)(−2.55 × 105 J)
(2)(9.81 m/s2)(250.0 m) −  
250.0 kg
vi = 4.
03
m2/
s2
+2.0
03
m2/
s2 = 6.
03
m2/
s2
90
×1
4×1
94
×1
vi = 83.3 m/s = 3.00 × 102 km/h
6. hi = 12.3 km
PEi + KEi = PEf + KEf
g = 9.81 m/s
KEf = PEi − PEf = mghi − mghf = mg∆h
KEi = 0 J
KEf = mg∆h = (0.1200 kg)(9.81 m/s2)(3.2 × 103 m) = 3.8 × 103 J
∆h = hi − hf = 3.2 km
PEf = mghf = mg(hi − ∆h) = (0.1200 kg)(9.81 m/s2)(12.3 × 103 m − 3.2 × 103 m)
2
PEf = (0.1200 kg)(9.81 m/s2)(9.1 × 103 m) = 1.1 × 104 J
Alternatively,
PEf = PEi − KEf = mghi − KEf
PEf = (0.1200 kg)(9.81 m/s2)(12.3 × 103 m) − 3.8 × 103 J = 1.45 × 104 J − 3.8 × 103 J
PEf = 1.07 × 104 J
7. h = 68.6 m
ME i = PE i = mgh
1
v = 35.6 m/s
MEf = KEf = 2 mv 2
g = 9.81 m/s2
mgh − 2 mv 2)
(MEi − MEf )(100)
percent of energy dissipated =  = 
(100)
MEi
mgh
PEf = 0 J
KEi = 0 J
1
1
gh − 2 v 2
percent of energy dissipated = 
(100)
gh
1
(9.81 m/s2)(68.6 m) − 2 (35.6 m/s)2
(100)
percent of energy dissipated = 
(9.81 m/s2)(68.6 m)
(673 J − 634 J)(100)
(39 J) (100)
percent of energy dissipated =  =  = 5.8 percent
673 J
673 J
II Ch. 5–8
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
PEi − PEf = KEf
m = 120.0 g