Question

a body of mass 5 kg is lifted vertically at a constant velocity of 12 m calculate the force applied to the work done in lifting the body what happens to work performed

Answer 1

Work = Force×Displacement.

Hence

W=F×S.

Here Force is not given.

According to Newton's second law of motion

Force=Mass×Acceleration (F=ma)

Mass=6 Kg

Acceleration (due to gravity)= 9.8 m/a.

Hence

F=6×9.8

F=58.8 N

Work= 58.8×12

Work Done=705.6 Joule

Hope this help...

Answer 2

$ANSWER$ $:-$

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.

$Hope It Helps U ..$

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