Download CJ Energy Assignment 4 Solutions

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CJ Energy Assignment 4
Work , Energy and Power 6.7 Power 55, 56, 59, 62 page192
POWER
𝑃=
𝑊𝑜𝑟𝑘
𝑊
=
𝑇𝑖𝑚𝑒
𝑡
Units of Power:
1 Joule/ second = 1 Watt
1 horse power = 746 Watts = 550 ft lbs/sec = 33,000 ft lbs/min
Text Questions
55. One Kilowatt-hour (kWh) is the amount of work or energy generated when one kilowatt of power is
supplied for a time of one hour. A kilowatt-hour is the unit of energy used by power companies when
determining your electric bill. Determine the number of Joules of Energy in one kilowatt-hour.
One kilowatt  hour is the amount of work or energy generated when one kilowatt of power is supplied for a time
3
of one hour. From Equation 6.10a, we know that W  P t . Using the fact that 1 kW = 1.0  10 J/s and that
1h = 3600 s, we have
1.0 kWh = (1.0  10 3 J/s)(1 h) = (1.0  103 J/s)(3600 s) = 3.6  10 6 J
56. Bicyclists in the Tour de France do enormous amounts of work during a race. For example, the average
power per kilogram generated by Lance Armstrong (m = 75.0 kg) is 6.50 W per kilogram of his body mass. (a)
How much work does he do during a 135-km race in which his average speed is 12.0 m/s? (b) Often, the work
done is expressed in nutritional calories rather than in joules. Express the work done in part (a) in terms of
nutritional calories, noting that 1 joule = 2.389 × 10−4 nutritional calories.
The work W done is equal to the average power P multiplied by the time t, or
W  Pt
(6.10a)
The average power is the average power generated per kilogram of body mass multiplied by Armstrong’s mass. The
time of the race is the distance s traveled divided by the average speed v , or t  s / v (see Equation 2.1).
SOLUTION
a. Substituting t  s / v into Equation 6.10a gives
 135 103 m 
W
s 
6
W  Pt  P     6.50
75.0
kg

  5.48 10 J

v
12.0
m/s
  


kg 

P

59. (Multiple Concepts) The cheetah is one of the fastest accelerating animals, because it can go from rest to
27 m/s (60 MPH) in 4 seconds. If its mass is 110 kg, determine the average power developed by the cheetah
during the acceleration. State your answer in (a) watts and (b) horsepower.
REASONING The average power developed by the cheetah is equal to the work done by the cheetah divided by the
elapsed time (Equation 6.10a). The work, on the other hand, can be related to the change in the cheetah’s kinetic energy
by the work-energy theorem, Equation 6.3.
SOLUTION
a. The average power is
P
W
t
(6.10a)
where W is the work done by the cheetah. This work is related to the change in the cheetah’s kinetic energy by
Equation 6.3, W  12 mvf2  12 mv02 , so the average power is
2
2
W 12 mvf  12 mv0
P

t
t
1 110 kg 27 m / s 2  1 110 kg 0 m / s 2


 2


 2
 1.0  104 W
4.0 s
b. The power, in units of horsepower (hp), is


 1 hp 
P  1.0  104 W 
  13 hp
 745.7 W 
62. A 1900-kg car experiences a combined force of air resistance and friction that has the same magnitude whether the
car goes up or down a hill at 27 m/s. Going up a hill, the car’s engine needs to produce 47 hp more power to sustain the
constant velocity than it does going down the same hill. At what angle is the hill inclined above the horizontal?
The following drawings show the free-body diagrams for the car in going both up and down the hill. The force FR is the
combined force of air resistance and friction, and the forces FU and FD are the forces supplied by the engine in going
uphill and downhill respectively.
Going up the hill
FN
Going down the hill
FN
FU
FR
FD
FR
mg cos 
mg sin 
mg sin 
mg cos 
Writing Newton's second law in the direction of motion for the car as it goes uphill, taking uphill as the positive
direction, we have
FU  FR  mg sin  = ma  0
Solving for FU , we have
FU  FR  mg sin 
Similarly, when the car is going downhill, Newton's second law in the direction of motion gives
FR  FD  mg sin  = ma  0
so that
FD  FR  mg sin 
Since the car needs 47 hp more to sustain the constant uphill velocity than the constant downhill velocity, we
can write
P U  P D  P
where P U is the power needed to sustain the constant uphill velocity, PD is the power needed to sustain the
constant downhill velocity, and P  47 hp . In terms of SI units,
746 W 
P   47 hp  
  3.51 10 4 W
 1 hp 
Using Equation 6.11 ( P  Fv ), the equation PU  PD  P can be written as
FU v  FD v  P
Using the expressions for FU and FD , we have
( FR  mg sin  )v  ( FR  mg sin  )v  P
Solving for , we find
 P
 2mgv
  sin –1 


3.51  104 W
–1 

sin

  2.0

 2(1900 kg)(9.80 m/s2 )(27 m/s) 



NON TEXT QUESTIONS
1. A sports car is said to have 300 horsepower, how many Watts is this?
300 _ horsepower
746 _ Watts

 223,800 _ Watts  223.8 _ Kilowatts
1
1 _ horsepower
2.
Lance Armstrong can put out 1,000 watts of power. How does this compare to horsepower? If another rider can put out only 800
watts, is he doomed to lose the race to Lance? (This is a logic question)
No, the other rider is not doomed to loose simply because Lance can produce more power. If the
other rider is lighter (has less mass) then his power to weight ratio is favored.
3.
A scientist studying the power of machines only has instruments to measure velocity of an object moved by a machine and the
force applied to create the velocity. Can the scientist calculate the watts output by the machine?

Power is defined as P  W . Recall that work is defined as W  F  x . Substitute the definition
t
of work into the definition of Power to get:
 
W F  x
P

t
t
 
F  x
P
t
Notice that within the new definition of power is the definition of velocity (Δx/Δt). This is equal
to Force x velocity.
4.
Person A moves a box across a level surface. The force required is 100 N and the box is moved 10 meters. Person A completes
this task in 100 seconds. What power is required?
 
F  x (100 N )(10m)
P

 10W
t
100 _ sec
5.
Person B moves a box across a level surface. The force required is 100 N and the box is moved 10 meters. Person B completes
this task in 50 seconds. What power is required?
 
F  x (100 N )(10m)
P

 20W
t
50 _ sec
6.
A 200 pound football player runs up the hill behind the football field (a height of 25 feet) in 7 seconds. What is the power of the
football player?
The football player is doing work against gravity and therefore increasing his Gravitational
Potential Energy.
From the Law of Conservation of Energy: W = E2 - E1
Since State 1 has no GPE the work done is equal to the energy in State 2, GPE2.
GPE2 = mgh
Substitute this into the equation for power.


m  g  h W  h (200lb)(25 ft )
P


 714 _ lb  ft / sec
t
t
7 _ sec
7.
A 100 pound football player runs up the hill behind the football field (a height of 25 feet) in 3.5 seconds. What is the power of the
football player?


m  g  h W  h (200lb)(25 ft )
P


 1,428 _ lb  ft / sec
t
t
3.5 _ sec
8.
At its top speed of 35 knots (65 km/hr) the aircraft carrier John F Kennedy experiences a friction force from the water of 11.6 x
106 Newtons. Assuming the engines are 100% efficient, what horsepower must the ships engines provide?
First convert the velocity to m/s.
65 _ km 1,000 _ m
1 _ hr


 18 _ m / s
hr
1 _ km 3,600 _ sec
 
F  x  
P
 F  v  (11.6  106 N )(18m / s )  20.8  108 J
t
9.
The space shuttle is by far the most powerful vehicle ever devised. The overall power of a space shuttle at takeoff is about 11 GW
or 11,000,000,000 watts of power. That is about 15 million horsepower! No calculation is necessary for this one, just say wow.
WOW