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(b) Work (W) is defined as a force (F) moved through a distance 3.19. The maximum velocity occurs at the lowest point with a gain of

(d), or W = Fd. Power (P) is defined as work (W) per unit kinetic energy equivalent to the loss of potential energy in

of time (t), or P = W/t. Th erefore, falling 3.0 in (which is 0.25 ft ), so
Fd _
P = KE gained = PE lost

t
1 _ 2
(170.0 lb)(25.0 ft)
__ mv = mgh
= 2
10.0 s
v = √ 2gh

__ _
(170.0)(25.0) ft·lb

= = √ (2)(32 ft/s 2 )(0.25 ft)

10.0 s

2
_ = √ (2)(32)(0.25) (ft/s × ft)

ft·lb
= 425
s 2
2

= √ 16 ft /s
_
ft·lb

One hp is defined as 550 and

s = 4.0 ft/s
_
425 ft·lb/s
= 0.77 hp 3.20. (a) W = Fd and the force F that is needed to lift the load

ft·lb/s

550 _ upward is mg, so W = mgh. Power is W/t, so

hp
mgh
_

Note that the student’s power rating (425 ft·lb/s) is less than the P =

power rating defined as 1 horsepower (550 ft ·lb/s). Th us, the t

2
(250.0 kg)(9.8 m/s )(80.0 m)
student’s horsepower must be less than 1 hp. A simple analysis such ___

=
as this will let you know if you inverted the ratio or not. 39.2 s
__ _ _ _ _
3.16. (a) The force (F) needed to lift the elevator is equal to the = (250.0)(9.8)(80.0) kg m m 1

× × ×

weight of the elevator. Since the work (W) is equal to 39.2 1 s 2 1 s
Fd and power (P) is equal to W/t, then 196,000 kg·m 2 1
_ _ _

= ×
Fd _ Fd _ 39.2 s 2 s
P = ∴ t =

t P J
_

= 5,000
(2,000.0 lb) (20.0 ft) s
__
= _ = 5.0 kW

( 550 ) (20.0 hp)
ft·lb
s _

hp (b) There is 746 watts per horsepower, so
_ _ 1 _ 5,000 W 5,000 hp
40,000 ft·lb

= × × hp _ _ W _ _

×

=

_
11,000 ft·lb hp W _ 746 1 W

s 746
hp
_ _ _
s
40,000 ft·lb

= × W·hp

_

11,000 1 ft·lb = 6.70
_ W
ft·lb·s
= 3.64 = 6.7 hp
ft·lb
= 3.6 s
− _
d
(b) v =

t CHAPTER 4
_
20.0 ft
=

3.6 s 4.1. List the known and unknown quantities:
= 5.6 ft/s
1 _ 2 body temperature T F = 98.6°
3.17. Since PE lost = KE gained , then mgh = mv . Solving for v,
2 T C = ?

2
v = √ 2gh = √ (2)(32.0 ft/s )(9.8 ft)

These are the quantities found in the equation for conversion

2
2

= √ (2)(32.0)(9.8) ft /s 5 _

of Fahrenheit to Celsius, T C = (T F – 32°), where T F is the

2
2

= √ 627 ft /s 9
= 25 ft/s temperature in Fahrenheit and T C is the temperature in Celsius.

1 _ 2 _ This equation describes a relationship between the two

2KE
3.18. KE = mv ∴ v = √
2 m temperature scales and is used to convert a Fahrenheit
temperature to Celsius. The equation is already solved for the

2 )
2
(
kg·m

(2) 200,000 _ Celsius temperature, T C . Th us,
s
__

= √

1,000.0 kg 5 _

T C = (T F – 32°)
9
2
400,000 kg·m
1

×

= _ _ _ 5 _

√ 1,000.0 s 2 kg = (98.6° – 32°)
2 9
kg·m
400.000 _

333°
= _ _
√ 1,000.0 kg·s 2 = 9
2
2
= √ 400 m /s = 37°C

= 20 m/s
E-11 APPENDIX E Solutions for Group A Parallel Exercises 653