Question

A man carries a load of 20N on his head over a horizontal distance of 20 m how much work is done by the man

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

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