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Two blocks are connected by a string as shown in figure. They are released from rest. Show that after they moved a distance L, their common speed in given by sqrt((2(m_(2)-mum_(1))gl)/((m_(1)+m_(2)))) where mu is the coefficient of friction

Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :In moving a distance L by `m_(1) and m_(2)` the <a href="https://interviewquestions.tuteehub.com/tag/pe-590719" style="font-weight:bold;" target="_blank" title="Click to know more about PE">PE</a> will be lost by `m_(2)` only and will be `m_(2)gl`, while work done by friction will be only on `m_(1)` and will be `<a href="https://interviewquestions.tuteehub.com/tag/mu-566056" style="font-weight:bold;" target="_blank" title="Click to know more about MU">MU</a> m_(1)gL`. Now as the string is inextensible `v_(1) = v_(2) = v ` and so, the gain in <a href="https://interviewquestions.tuteehub.com/tag/ke-527890" style="font-weight:bold;" target="_blank" title="Click to know more about KE">KE</a> wil be = `(1//2) (m_(1) + m_(2)) v^(2)`. By conservation of energy. <br/> Loss in PE = Grain in KE + <a href="https://interviewquestions.tuteehub.com/tag/wd-729586" style="font-weight:bold;" target="_blank" title="Click to know more about WD">WD</a> against friction <br/> i.e., `m_(2)gL =(1)/(2)(m_(1)+m_(2))v^(2)+mu m_(1)gL` <br/> `v=sqrt((2(m_(2)-mum_(1))gL)/((m_(1)+m_(2))))`</body></html>


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