JEE Advance - Physics (2009 - Paper 2 Offline - No. 12)
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/s2, 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.
Giải thích
We have,
$$a = \left( {{{{m_1} - {m_2}} \over {{m_1} + {m_2}}}} \right)g = \left( {{{0.72 - 0.36} \over {0.72 - 0.36}}} \right) \times 10 = {g \over 3} = {{10} \over 3}$$
$$T = {{2{m_1}{m_2}g} \over {{m_1} + {m_2}}} = {{2 \times 0.72 \times 0.36 \times 10} \over {0.72 + 0.36}} = 4.8$$ N
$$s = {1 \over 2}a{t^2} = {1 \over 2} \times {{10} \over 3} \times {1^2} = {5 \over 3}$$ m
The work done by the rope on 0.36 kg is
$$W = Ts\cos 0^\circ = 4.8 \times {5 \over 3} = + 8$$ J