Work, Power and EnergyHard
Question
A 1000 kg elevator rises from rest in the basement to the fourth floor, a distance of 20 m. As it passes the fourth floor its speed is 4m/sec. There is a constant frictional force of 500 N. The work done by the lifting mechanism is:-
Options
A.196 × 103 J
B.204 × 103 J
C.214 × 103 J
D.203 × 105 J
Solution
Work done against gravitational force = mgh
= 1000 × 9.8 × 20 = 196 × 103 J
Work done to impart velocity to the body
=
mV2
=103 × 16 = 8 × 103J
Work done against frictional force
= 500 × 20 = 10 × 103 J
Total work done = 214 × 103 J
= 1000 × 9.8 × 20 = 196 × 103 J
Work done to impart velocity to the body
=
mV2=
103 × 16 = 8 × 103JWork done against frictional force
= 500 × 20 = 10 × 103 J
Total work done = 214 × 103 J
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