Question

1. What is the work done by a man in carrying a suitcase weighing 30 kg over his head, when he travels a distance of 10 m in the (i) vertical direction (ii) horizontal direction.
(answer should come 2,940 Joules)

2) 230 joule were spent in lifting a 10 kg weight to a height of 2 m. Calculate the acceleration with which it was raised. Take g = 10 m$s^{-2}$.
(answer should come 1.5m$s^{-2}$)

3) A block of mass 2 kg is lying on the frictionless table. A force of 8 N is applied on it for 12 s. Calculate its kinetic energy. (answer should come 2,304 Joules)

Answer 1

Work done in horizontal direction = 0 as the potential energy is not changed and the change in kinetic energies : initial and final - is zero.

work done when the person carries the suitcase in vertical direction =
   = m g h = change in potential energy = 30 kg * 9.8 m/sec/sec * 10 m
   =  2940 Joules
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There is a force F1 lifting the body upwards. There is Weight W = m g pulling the body downwards.  The net force F2 on the body is upwards with acceleration a.

   F2 = m a  = F1 - W = F1 - m g
           a =  F1 / m - g

Work done by Force F1 with which the object is lifted = F1 . s = 230 Joules
      F1  = energy spent / distance = 230 J / 2 m  = 115 Newtons

Net Acceleration of the body = a = F1 / m - g = 115/10 - 10  = 1.5 m/sec/sec
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 m = 2 kg          F = 8 Newtons        t = 12 sec
 a = F/m = 4 m/sec/sec
v = u + a t = 0 + 4 * 12 = 48 m/sec

kinetic energy = 1/2 m v² = 1/2 * 2 * 48 * 48 = 48² Joules