Two blocks of masses M and m are placed on on | vedclass

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(C) Since the blocks move together with no relative sliding,the acceleration $a$ of the system is given by Newton's second law: $a = \frac{F}{M+m}$.Th

Vedclass · Accepted Answer

$\frac{m F^2 t^2}{2(M+m)^2}$

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(C) Since the blocks move together with no relative sliding,the acceleration $a$ of the system is given by Newton's second law: $a = \frac{F}{M+m}$.The friction force $f$ acting on the upper block of mass $m$ provides the necessary acceleration for it to move with the lower block. Thus,$f = ma = m \left( \frac{F}{M+m} ight) = \frac{mF}{M+m}$.The distance $s$ covered by the blocks in time $t$ starting from rest is $s = \frac{1}{2} a t^2 = \frac{1}{2} \left( \frac{F}{M+m} ight) t^2$.The work done by friction on the upper block is $W = f 	imes s = \left( \frac{mF}{M+m} ight) 	imes \left( \frac{1}{2} \frac{F t^2}{M+m} ight) = \frac{m F^2 t^2}{2(M+m)^2}$.Note: The work done by friction on the lower block is equal in magnitude but opposite in sign $(-W)$,so the net work done by friction on the system is zero. The question asks for the work done by friction on the blocks (implying the upper block),which is $\frac{m F^2 t^2}{2(M+m)^2}$.

Two blocks of masses $M$ and $m$ are placed on one another on a smooth horizontal surface as shown in the figure. The force $F$ is acting on the mass $M$ horizontally during time interval $t$. Assuming no relative sliding between the blocks,the work done by friction on the blocks is .......... .

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