A person lifts a 25 kg block hanging over a fixed light frictionless small pulle

Question

A person lifts a 25 kg block hanging over a fixed light frictionless small pulley by walking horizontally, as shown in figure. As the person walks 2 metres, the angle of the rope to the horizontal changes from 60 degree to 30 degree. If the block rises at constant speed. the work done by rope on the person as the person moves by 2 metres is: [consider the rope to be light and inextensible] (Take g = 10m/s^2) (A) 500 (squareroot 3 - 1)J (B) 500(1 - squareroot 3)J (C) 500 (squareroot 3)J (D) None of these

Explanation / Answer

Tension in the rope, T = mg = 25 x 10 = 250 N

Horizontal component of this tension on the other side of the pulley

Tx = T*cos theta

average force, Fav = (T*cos30 + T*cos50) / 2 = 250*(cos30 + cos50) / 2

= 125*(0.866 + 0.643) = 188.63 N

So, work done by the person = Fav*d = 188.63*2 = 377.25 J

So, work done by the rope on the person = (Negative)work done by the person = - 377.25 J

This value is closer to option (B).

So, option (B) is the correct answer.

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