Search any question & find its solution
Question:
Answered & Verified by Expert
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 .......... .
Options:
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 .......... .
Solution:
2941 Upvotes
Verified Answer
The correct answer is:
$\frac{m F^2 t^2}{2(M+m)^2}$
From (i) \& (ii), we get $a=\frac{F}{m+M}$ and $f=\frac{m F}{m+M}$ Distances moved in time $t(s)$
$$
=\frac{1}{2} a t^2=\frac{1}{2}\left(\frac{F}{M+m}\right) t^2
$$
Work done by friction $=f \times s$
$$
=\left(\frac{m F}{M+m}\right) \times \frac{1}{2} \frac{F t^2}{(M+m)}=\frac{m F^2 t^2}{2(M+m)^2}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.