Work:W
= F x d = Fd cosq

1.How
much work does Fred do exerting 45 N to lift a box 3.2 m?
W
= F x W
= (45)(3.2) W
= 144 J 
2.What
distance will 90 J of energy slide a box against 12 N of frictional force?
W
= F x d 90
= (12)d 90s
= 12 d = 90/12 d = 7.5 m 
3.Katherine
moves a box 7.2 m doing 9 J of work.What
is the fricitonal force?
W
= F x d 9
= F(7.2) F
= 9/7.2 F
= 1.25 N 
Work
and weight:DE_{p
}= mgDh

4.How
much work is needed to lift a 45 Kg box 9.2 m?
DE_{p
}= mgDh DE_{p
}= (45)(9.8)(9.2) DE_{p
}= 4057.2 DE_{p
}= 4060 J 
5.A
pump does 4500 J of work in a minute.What
mass of water can it pump to a height of 4.5 m in one minute?
DE_{p
}= mgDh 4500
= m(9.8)(4.5) m
= 4500/(9.8)(4.5) m
= 102 Kg 
6.A
rifle cartridge contains 250 J of energy.How
far can it launch its 60 gram slug into the air if it is shot straight
up?
DE_{p
}= mgDh 250
= (60 g)(9.8)h 250
= (.06 Kg)(9.8)h 250
= (.588)h h
= 250/.588 h
= 425 m 
Power:P
= DW/Dt

7.A
motor can do 540 J of work in 9.0 seconds.What's
its power output?
P
= DW/Dt P
= 540/9 P
= 60 W 
8.In
what time can a 250 W motor do 4500 J of work?
P
= DW/Dt 250
= 4500/Dt Dt
= 4500/250 Dt
= 18 s 
9.How
much work does a 25 W motor do in 5 minutes?
P
= DW/Dt 25
= DW/5
min 25
= DW/(5
min)(60 s/1 min) 25
= DW/300 DW
= (25)(300) DW
= 7500 J 
Jambalaya:W
= F x s,DE_{p
}= mgDh,P
= DW/Dt

10.What
must be the power rating of a motor that can lift a 600 Kg load 30 meters
in 5 seconds?
DE_{p
}= mgDh DE_{p
}= (600)(9.8)(30) DE_{p
}= 176400 DE_{p
}= DW P
= DW/Dt P
= 176400/5 P
= 35280 W 
11.A
sled dog drags a 240 Kg sled 50 m in 35 seconds when the coefficient of
friction between the snow and the runners is .056.What
is the power output of the dog?
W
= F x d,P = DW/Dt,F_{fr}£m_{k}F_{N},F_{N}
= mg F_{fr}£m_{k}F_{N}F_{N}
= mg F_{fr}£
(.056)(240)(9.8) F_{fr}£
131.712 W
= F x d W
= (131.712)(50) W
= 6565.6 P
= DW/Dt P
= 6585.6/35 P
= 188 W 
12.Greg
LeMond can put out ¾ of a horsepower.(1
HP = 745.7 W).In what time can he
climb a 2000 m mountain if he and his bike have a mass of 82 Kg? (Ignore
Friction)
DE_{p
}= mgDh DE_{p
}= (82)(9.8)(2000) DE_{p
}= 1607200 J DE_{p
}= DW (3/4)(745.7)
= 559.275 W P
= DW/Dt 560
= 1607200/Dt Dt
= 1607200/560 Dt
= 2870 s 
13.My
van can only go about 25 m in one second on a level road at full power.(30
HP,1 HP = 745.7 W)What must be the
force of friction opposing my van?(most
of this is air friciton and drag)
W
= F x s,P = DW/Dt P
= (30)(745.7) = 22371 W P
= DW/Dt 22371
= DW/1 W
= F x d 22371
= F(25) F
= 900 N 
14.A
steam engine must drag logs across 45 m of level rock.if
the coefficient of friction between the logs and the ground is .78, and
the engine can put out 5 Horse power (1 HP = 745.7 W), what is the maximum
mass of logs it can drag in 15 seconds?
W
= F x d,P = DW/Dt,F_{fr}£m_{k}F_{N},F_{N}
= mg P
= (5)(745.7) = 3728.5 3728.5
= DW/15 DW
= (3728.5)(15) DW
= 55927.5 W
= F x d 55927.5
= F(45) F
= 55927.5/45 F
= 1241.8 F_{fr}£m_{k}F_{N},F_{N}
= mg 1241.8 £
(.78)(9.8)(m) m
= 163 kg 
15.A
hiker can put out 250 W of power.If
they have a mass of 72 Kg and are carrying a 43 Kg pack, what elevation
gain can they expect to achieve in 6 hours of hiking?
P
= DW/Dt DE_{p
}= mgDh DE_{p
}= DW P
= mgDh/Dt 250
= (72 + 43)(9.8)h/6 hr 250
= (115)(9.8)h/(6 hr)(60 min/1 hr)(60 s/1 min) 250
= (1127)h/21600 s 5400000
= 1127h h
= 5.4 x 10^{6}/1127 h
= 4791.48 h
= 4800 m 