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. [2009]

Question

Smart Achievers · Accepted Answer

(8)

When the system is released,

$T - m g = m a ...(i)$

$M g - T = M a ...(ii)$

$From eq. (i) & (ii)$

$a = \frac{(M - m) g}{M + m} = \frac{g}{3}$

$and T = \frac{4 m g}{3}$

$For block m = 0.36 kg$

$u = 0, a = \frac{g}{3}, t = 1, s = ?$

$s = u t + \frac{1}{2} a t^{2} = 0 + \frac{1}{2} \times \frac{g}{3} \times 1^{2} = \frac{g}{6} (m = 0.72 kg)$

$∴$ $Work done by the string on m$

$W = T s \cos 0 ° = \frac{4 m g}{3} \times \frac{g}{6} \times 1 = \frac{4 \times 0.36 \times 10 \times 10}{3 \times 6} = 8 J$

Solution

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Solution

Ans. (8) When the system is released, T-mg=ma ...(i) Mg-T=Ma ...(ii) From eq. (i) & (ii) a=(M-m)gM+m=g3 and T=4mg3 For block m=0.36 kg u=0, a=g3, t=1, s=? s=ut+12at2=0+12×g3×12=g6 (m=0.72 kg) ∴ Work done by the string on m W=Tscos0°=4mg3×g6×1=4×0.36×10×103×6=8 J

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