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A block of mass m is pulled along a horizontal surface by applying a force at an angle theta with the horizontal. If the block travels with a uniform velocity and has a displacement d and the coefficient of friction is mu, then the work done by the applied force is

Answer» <html><body><p>`(mu mgd)/(cos theta + mu sin theta)` <br/>`(mu m g d cos theta)/(cos theta + mu cos theta)` <br/>`(mu m g d sin theta)/(cos theta + mu sin theta)`<br/>`(mu m g d cos theta)/(cos theta - mu sin theta)` </p>Solution :<img src="https://doubtnut-static.s.llnwi.net/static/physics_images/MTG_NEET_GID_PHY_XI_C04_E01_001_S01.png" width="80%"/> <br/> Because the block moves with a uniform velocity, the resultant force is zero. Resolving <a href="https://interviewquestions.tuteehub.com/tag/f-455800" style="font-weight:bold;" target="_blank" title="Click to know more about F">F</a> into horizontal components `F cos theta` and vertical component `F sin theta,` we <a href="https://interviewquestions.tuteehub.com/tag/get-11812" style="font-weight:bold;" target="_blank" title="Click to know more about GET">GET</a> <br/> `R + F sin theta = mg orR = mg - F sin theta` <br/> Also , `f = mu R = mu(mg - F sin theta)` <br/> But `f = F cos theta` <br/> `:. F cos theta = mu (mg - F sin theta)` or `F (cos theta + mu sin theta) = mumg` <br/> or `F = (mu mg)/(cos theta + mu sin theta theta)` <br/> work done, `<a href="https://interviewquestions.tuteehub.com/tag/w-729065" style="font-weight:bold;" target="_blank" title="Click to know more about W">W</a> = <a href="https://interviewquestions.tuteehub.com/tag/fs-1000993" style="font-weight:bold;" target="_blank" title="Click to know more about FS">FS</a> cos theta, W= (mu m g d cos theta)/(cos theta + mu sin theta) ( :. s = d)`</body></html>


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