A body of mass 5 kg is lifted vetically at a constant velocity of 12m. Calculate-

1.force applied.

2.work done in lifting the body.

3.what happens to the work performed?


The force applied on the body is equal to its weight = mg     (as body moving with constant velocity)
1) force applied = mg = 5*10 = 50 N
2) Net work done on the body is zero as net force is zero.
     
3)Work done by lifting force is cancelled by work done by gravitaional force.

  • 2

Here minimum force required to raise is mg

F=mg

F=5x10=50N

KE=1/2mv2

KE=1/2x5x12x12

KE=72x5

KE=360 joules

  • -3
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