Question bank/JEE Main/Physics

A block of mass 10 kg is to be pulled on a horizontal rough
surface (µ =0.75) the minimum magnitude of applied force is

A75 NB80 NC60 NDNone of these

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$Fcos\theta - \mu N = 0$
$N = mg - {Fsin}\theta\ldots\ldots(2)$
From (1) and (2)
${Fcos}\theta = \mu\left( mg - {Fsin}\theta \right)$
$K = \frac{\mu mg}{cos\theta + \mu sin\theta}$
$F_{\min} = \frac{\mu mg}{\sqrt{1 + \mu^{2}}}\left( \because y = Asin\theta + Bcos\theta;ymax = \sqrt{A^{2} + B^{2}} \right)$
$F_{\min} = \frac{0.75 \times 10 \times 10}{\sqrt{1 + (0.75)^{2}}}$

Source: Grass-G / P-06 Physics paper 14 October.docx