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Holt Physics
Problem 5B
KINETIC ENERGY
PROBLEM
Silvana Cruciata from Italy set a record in one-hour running by running
18.084 km in 1.000 h. If Cruciata’s kinetic energy was 694 J, what was her
mass?
SOLUTION
1. DEFINE
2. PLAN
Given:
∆x = 18.084 km = 1.8084 × 104 m
∆t = 1.000 h = 3.600 × 103 s
KE = 694 J
Unknown:
m=?
Choose the equation(s) or situation: Use the definition of average velocity to
calculate Cruciata’s speed.
∆x
vavg = 
∆t
Use the equation for kinetic energy, using vavg for the velocity term, to solve for
m.
1
KE =  m vavg2
2
2KE
2KE 2 2KE∆t 2
m = 2 = ∆x = 

vavg
∆x 2
∆t
3. CALCULATE
Substitute the values into the equation(s) and solve:
(2)(694 J)(3.600 × 103 s)2
m = 
= 55.0 kg
(1.8084 × 104 m)2
4. EVALUATE
If the average speed is rounded to 5.0 m/s, and the kinetic energy is rounded to
700 J, the estimated mass is 56 kg, which is close to the calculated value.
ADDITIONAL PRACTICE
1. In 1994, Leroy Burrell of the United States set what was then a new world
record for the men’s 100 m run. He ran the 1.00 × 102 m distance in 9.85 s.
Assuming that he ran with a constant speed equal to his average speed, and
his kinetic energy was 3.40 × 103 J, what was Burrell’s mass?
2. The fastest helicopter, the Westland Lynx, has a top speed of 4.00 ×
102 km/h. If its kinetic energy at this speed is 2.10 × 107 J, what is the
helicopter’s mass?
42
Holt Physics Problem Workbook
Copyright © by Holt, Rinehart and Winston. All rights reserved.
Rearrange the equation(s) to isolate the unknown(s): Substitute the average
velocity equation into the equation for kinetic energy and solve for m.
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3. Dan Jansen of the United States won a speed-skating competition at the
1994 Winter Olympics in Lillehammer, Norway. He did this by skating
500 m with an average speed of 50.3 km/h. If his kinetic energy was
6.54 × 103 J, what was his mass?
4. In 1987, the fastest auto race in the United States was the Busch Clash in
Daytona, Florida. That year, the winner’s average speed was about
318 km/h. Suppose the kinetic energy of the winning car was 3.80 MJ.
What was the mass of the car and its driver?
5. In 1995, Karine Dubouchet of France reached a record speed in downhill
skiing. If Dubouchet’s mass was 51.0 kg, her kinetic energy would have
been 9.96 × 104 J. What was her speed?
6. Susie Maroney from Australia set a women’s record in long-distance
swimming by swimming 93.625 km in 24.00 h.
a. What was Maroney’s average speed?
b. If Maroney’s mass was 55 kg, what was her kinetic energy?
7. The brightest, hottest, and most massive stars are the brilliant blue stars
designated as spectral class O. If a class O star with a mass of 3.38 × 1031 kg
has a kinetic energy of 1.10 × 1042 J, what is its speed? Express your answer
in km/s (a typical unit for describing the speed of stars).
8. The male polar bear is the largest land-going predator. Its height when
standing on its hind legs is over 3 m and its mass, which is usually
around 500 kg, can be as large as 680 kg. In spite of this bulk, a running
polar bear can reach speeds of 56.0 km/h.
a. Determine the kinetic energy of a running polar bear, using the
maximum values for its mass and speed.
Copyright © by Holt, Rinehart and Winston. All rights reserved.
b. What is the ratio of the polar bear’s kinetic energy to the kinetic
energy of Leroy Burrell, as given in item 1?
9. Escape speed is the speed required for an object to leave Earth’s orbit. It
is also the minimum speed an incoming object must have to avoid being
captured and pulled into an orbit around Earth. The escape speed for a
projectile launched from Earth’s surface is 11.2 km/s. Suppose a meteor
is pulled toward Earth’s surface and, as a meteorite, strikes the ground
with a speed equal to this escape speed. If the meteorite has a diameter of
about 3 m and a mass of 2.3 × 105 kg, what is its kinetic energy at the
instant it collides with Earth’s surface?
Problem 5B
43
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Additional Practice 5B
Givens
Solutions
1. ∆x = 1.00 × 102 m
∆t = 9.85 s
3
KE = 3.40 × 10 J
∆x
v = 
∆t
2
∆x
1
1
KE = 2mv 2 = 2m 
∆t
2KE∆t2 (2)(3.40 × 103 J)(9.85 s)2
m = 
= 
= 66.0 kg
(1.00 × 102 m)2
∆ x2
2. v = 4.00 × 102 km/h
KE = 2.10 × 107 J
3. v = 50.3 km/h
3
KE = 6.54 × 10 J
4. v = 318 km/h
(2)(6.54 × 103 J)
2KE
= 67.0 kg
m = 
2 = 
(50.3 km/h)2(103 m/km)2 (1 h/3600 s)
v
(2)(3.80 × 106 J)
2KE
m = 
= 974 kg
2 = 
(318 km/h)2 (103 m/km)2 (1 h/3600 s)
v
KE = 3.80 MJ
5. m = 51.0 kg
× 10 J)
2mKE = (2
)(95.916.0
=
kg
II
4
4
KE = 9.96 × 10 J
v=
62.5 m/s = 225 km/h
∆t = 24.00 h
∆x
9.3625 × 104 m
a. vavg =  =  = 1.084 m/s
∆t
(24.00 h)(3600 s/h)
m = 55 kg
b. KE = 2mv 2 = 2(55 kg)(1.084 m/s)2 =
6. ∆x = 93.625 km
1
7. m = 3.38 × 1031 kg
42
KE = 1.10 × 10 J
Copyright © by Holt, Rinehart and Winston. All rights reserved.
(2)(2.10 × 107 J)
2KE

=
= 3.40 × 103 kg
m = 
2
(4.00 × 10 km/h)2 (103 m km)2 (1 h/3600 s)2
v2
8. m = 680 kg
v = 56.0 km/h
KELB = 3.40 × 103 J
9. v = 11.2 km/s
1
× 10 J)
= 2.55 × 10 m/s =
2mKE = (2
3).(318.1×01
0kg
42
v=
32 J
31
1
5
255 km/s
1
a. KE = 2mv 2 = 2(680 kg)[(56.0 km/h)(103 m/km)(1 h/3600 s)]2 = 8.23 × 104 J
24
KEp
8.2 × 104 J
b. b = 
= 
1
KELB 3.40 × 103 J
1
1
KE = 2mv 2 = 2(2.3 × 105 kg)(11.2 × 103 m/s)2 = 1.4 × 1013 J
m = 2.3 × 105 kg
Section Two — Problem Workbook Solutions
II Ch. 5–3