IIT-JEE 2009 · Paper 2 · Physics Q55

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PhysicsInteger+4 / -1Question 55 of 57

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 $g=10\ \mathrm{m/s^2}$ , 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.

A fixed pulley hanging from a support, with a small block (0.36 kg) on the left end of the string and a larger block (0.72 kg) on the right end.

Your answer

Answer and worked solution

Official answer: 8

Atwood machine. $a=\dfrac{(0.72-0.36)g}{1.08}=\dfrac{10}3\ \mathrm{m/s^2}$ ; tension $T=0.36(g+a)=0.36\times\dfrac{40}3=4.8$ N.

Distance in 1 s. $s=\frac12at^2=\frac53$ m, upward for the lighter block.

Work by the string. $W=Ts=4.8\times\frac53=8$ J.

Answer: 8

Topics:Constraint motion and pulleys Work by constant and variable forces

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IIT-JEE 2009 · Paper 2, Physics question 55