AP EAMCET · PHYSICS · Laws of Motion
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 .......... .
- A \(\frac{F t}{2(M+m)}\)
- B \(\frac{M+m}{m t^2}\)
- C \(\frac{m F^2 t^2}{2(M+m)^2}\)
- D \(\frac{F^2 t^2}{(M+m)}\)
Answer & Solution
Correct Answer
(C) \(\frac{m F^2 t^2}{2(M+m)^2}\)
Step-by-step Solution
Detailed explanation
For ' \(m\) ' mass \(f=ma\) ...(i) For ' \(M\) ' mass \(F-f=m a\) ...(ii) 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…
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