JEE MAIN - Physics (2023 - 1st February Evening Shift - No. 11)
As shown in the figure a block of mass 10 kg lying on a horizontal surface is pulled by a force F acting at an angle $$30^\circ$$, with horizontal. For $$\mu_s=0.25$$, the block will just start to move for the value of F : [Given $$g=10~\mathrm{ms}^{-2}$$]
_1st_February_Evening_Shift_en_11_1.png)
25.2 N
35.7 N
20 N
33.3 N
Explanation
_1st_February_Evening_Shift_en_11_2.png)
$$F\cos 30^\circ = {\mu _s}N$$ ........(1)
$ F \sin 30^{\circ}+\mathrm{N}=\mathrm{mg} $
$\Rightarrow \mathrm{N}=\mathrm{mg}-\mathrm{F} \sin 30^{\circ} $ ........(2)
From equation 1,
$\mathrm{F} \cos 30^{\circ}=\mu_{\mathrm{s}}\left(\mathrm{mg}-\mathrm{F} \sin 30^{\circ}\right) $
$ \mathrm{F} \cos 30^{\circ}=\mu_{\mathrm{s}} \mathrm{mg}-\mu_{\mathrm{s}} \mathrm{F} \sin 30^{\circ} $
$ \mathrm{F}\left(\cos 30^{\circ}+\mu_{\mathrm{s}} \sin 30^{\circ}\right)=\mu_{\mathrm{s}} \mathrm{mg} $
$ \mathrm{F}=\frac{\mu_{\mathrm{s}} \mathrm{mg}}{\cos 30^{\circ}+\mu_{\mathrm{s}} \sin 30^{\circ}}=\frac{0.25 \times 10 \times 10}{\sqrt{3} / 2 \times 0.25 \times 1 / 2} $
$$ \Rightarrow $$ $ \mathrm{~F}=\frac{25}{\sqrt{3} / 2+\frac{0.25}{2}}=\frac{50}{1.73+0.25}=\frac{50}{1.98}=25.2 \mathrm{~N}$
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