A block of mass 10 kg is placed on a horizon | vedclass

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(B) Let $M = 10 \ kg$ be the mass on the horizontal surface and $m$ be the mass of the hanging block.Since the system is released from rest,the initia

Vedclass · Accepted Answer

$11.11$

Vedclass · Answer

(B) Let $M = 10 \ kg$ be the mass on the horizontal surface and $m$ be the mass of the hanging block.Since the system is released from rest,the initial velocity $u = 0$.The distance covered by the hanging block is $s = 2 \ m$ in time $t = 2 \ s$.Using the equation of motion $s = ut + \frac{1}{2}at^2$:$2 = 0 + \frac{1}{2} 	imes a 	imes (2)^2$$2 = 2a \Rightarrow a = 1 \ ms^{-2}$.Now,applying Newton's second law to the system:For the hanging block: $mg - T = ma$For the block on the surface: $T = Ma$Adding these two equations: $mg = (M + m)a$$m(g - a) = Ma$$m(10 - 1) = 10 	imes 1$$9m = 10 \Rightarrow m = \frac{10}{9} \ kg \approx 1.11 \ kg$.The weight of the hanging block is $W = mg = \frac{10}{9} 	imes 10 = \frac{100}{9} \ N \approx 11.11 \ N$.

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