Question

A monkey of mass 60 kg climbs up a slippery pole with an acceleration of 2m/s^2for 3 seconds and subsequently slips down and comes to the starting position.FIND THE WORK DONE BY THE MONKEY DURING UPWARD MOTION.

Answer 1

At the maximum height, the final velocity will be ZERO.

Vf = 0

Vi = ?

a = 2 m/s^2

t = 3 seconds

Now using Equation of Motion, we have

Vf = Vi + at

0 = Vi + (2)(3)

Vi = -6 m/s (in upward direction)

Vi = 6 m/s

Now again using Equation of Motion, we have

2aS = (Vf)^2 - (Vi)^2

2(3)S = 0 - 36

6S = - 36

S = - 6 m (in upward direction)

S = h = 6 m

Now

Work done = mgh

                   =(60)(10)(6)

                   = 3600 N