Work Energy and Power

WORK, ENERGY AND POWER 1. An elevator with a 10-person capacity (each person with an average mass of 68 kg) raises passengers to a height of152 m in 55.0 s at constant speed. Find the power output of the motor assuming no friction losses?

P=

FS

=

( 68∗9.8∗10 ) ( 152 )

t

55

=18,416.87 s

joule

=18,416.87 watts

Ans. 18,416.87 watts

2. A child pulls a 5.6 kg box a distance of 12 m along a horizontal surface at a constant speed. What work does the child do on the box if the coefficient of kinetic friction is 0.20 and the cord makes an angle of 45° with the horizontal? Solution:

∑

F x =0 f −Fcos 45=0 μk N − Fcos 45=0 0.2 N −Fcos 45=0 Fcos 45 N = 0.2

∑

F y =0 N + Fsin 45−W =0 N + Fsin 45−5.6( 9.8)=0 N + Fsin 45−54.88= 0 1

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Fcos 45

+ Fsin 45−54.88=0 0.2 F=12.935 N W =FS=12.935 (12)=155.224 joule W =μk

NS =

μk∗Fcos 45

N

∗12=109.76

0.2 Ans.155.224 joule

3. A horse pulls a cart with a force of 42.0 lbs at an angle of 27° with the horizontal and moves along at a speed of 6.20

ml hr

.

a) How much work does the horse do in 12 min? b) Find he power output of the in hp Solution:

v =6.2

1m s

ml

hr 2.23 mi hr

=2.78

m

x

60 s

s 1 min

x

1 ft

=547.297 ft /min 0.3048 m

ft S=vt = 547.297 (12 min)=6567.565 ft min W =Fcos 27∗S=(42 cos 27 )(6567.565)=245,773.206lbs−ft lbs− ft P=Fvcos 27=42 (547.297) cos 27=20,481.1005 min lbs−ft 1 hp P=20,481.1005 x =0.62hp min lbs−ft 33000 min

( )

2

a) 245,773.206 lbs−ft b) 0.62 hp

Ans.

4. A body of mass m starts down from the top of an inclined plane 20ft long and 10 ft high. What is its velocity at a point 12 ft from the top if coefficient of friction is 0.1? Solution:

y 10 20−12 = 20 y=4 ft

1 21

m v2+ mgh = 1

1

mv 2 +mgh +W

1

2 f 21 2 m v2+ mgh = mv 2 +mgh + mgμ

1 1 2 2 2 1 2 gh = v + gh + gμ

2

1

2 2 2 1 2 (32.2) (10 )= v +32.2( 4)+(32.2)(0.1)

22 v =379.9 f t 2 6 2 s ft v =19.49 2 2

2

s

Ans.

v =19.49 2

3

ft s

5. To push a 25 kg crate up a 27° incline, a worker exerts a force of 120 N, parallel to the incline. As the crate slides 3.6 m, how much is done on the crate by each of the following: a) the worker b) the force of gravity, and c) the normal force due to the incline? Solution: a) W =FDcosθ=120( 3.6 )cos 0=432 joules b) W =FDcosθ=(25∗9.8)(3.6 )cos (27+90)=−4000.42 joules

c) W =FDcosθ=120(3.6 )cos 90=0 joule

Ans. a) 432 Joules b) 400.42 Joule c) 0 Joule

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