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

Answer:

3528 N

Step by step explanation:

It is given that the mass of the monkey is 60 kg and acceleration $a=2m/s^2$. After the monkey reaches the top, its final velocity $V_f=0.$

In order to get the work done, we first need to calculate the initial velocity $V_i.$

Using Newton's law of motion:

$V_f=V_i+at\\0=V_i+2(3)\\V_i=-6m/s=6m/s$ in the upward direction.

Use the equation of motion again to get the distance. Therefore,

$2ad=V_f^2-V_i^2\\2(3)d=6^2-0\\d=6\:m$

Work done = m*g*h; where h is the distance and g is the gravitational force of acceleration ($9.8\:m/s^2$)

$W=60*9.8*6=3528\:N$