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BITSAT 2013 Paper

Section: Physics

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Question : 9 of 150

Marks: +1, -0

A light inextensible string that goes over a smooth fixed pulley as shown in the figure connects two blocks of masses 0.36 kg and 0.72 kg. Taking g = 10 m/

s^2

, find the work done (in joules) by the string on the block of mass 0.36 kg during the first second after the system is released from rest.

4 J
2 J
8 J
10 J

Solution:

Given m = 0.36 kg, M = 0.72 kg.

The figure shows the forces on m and M.

When the system is released, let the acceleration be a. Then

T - mg = ma

Mg - T = Ma

∴ a =

\frac{(M-m)g}{M+m}

= g/3

and T = 4 mg/3

For block m :

u = 0 , a = g/3 , t = 1 , s = ?

s = ut +

\frac{1}{2} a t^2

= 0 +

\frac{1}{2} \times \frac{g}{38} 1^2

= g/6

∴ Work done by the string on m is

\overset{\rightarrow}{\overset{\Rightarrow}{\mathrm{T}}}

=

\frac{4 mg}{3} \times \frac{g}{6}

=

\frac{4 \times 0.36 \times 10 \times 10}{3 \times 6}

= 8J

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