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QJEEM23Ph808f: JEE Main 2023 Physics - Class 11 JEEM23Ph808f Solution: As shown in the figure a block of mass 10 kg lying on a…

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∘30^\circ , with horizontal. For μs=0.25\mu_s=0.25 , the block will just start to move for the value of F : [Given g=10 ms−2g=10~\mathrm{ms}^{-2} ]

JEE Main 2023 (Online) 1st February Evening Shift Physics - Laws of Motion Question 41 English

Multiple Choice Options

  • A 25.2 N
  • B 35.7 N
  • C 20 N
  • D 33.3 N

Step-by-Step Verified Solution

JEE Main 2023 (Online) 1st February Evening Shift Physics - Laws of Motion Question 41 English Explanation

Fcos⁡30∘=μsNF\cos 30^\circ = {\mu _s}N ........(1)

Fsin⁡30∘+N=mgF \sin 30^{\circ}+\mathrm{N}=\mathrm{mg}

⇒N=mg−Fsin⁡30∘\Rightarrow \mathrm{N}=\mathrm{mg}-\mathrm{F} \sin 30^{\circ} ........(2)

From equation 1,

Fcos⁡30∘=μs(mg−Fsin⁡30∘)\mathrm{F} \cos 30^{\circ}=\mu_{\mathrm{s}}\left(\mathrm{mg}-\mathrm{F} \sin 30^{\circ}\right)

Fcos⁡30∘=μsmg−μsFsin⁡30∘\mathrm{F} \cos 30^{\circ}=\mu_{\mathrm{s}} \mathrm{mg}-\mu_{\mathrm{s}} \mathrm{F} \sin 30^{\circ}

F(cos⁡30∘+μssin⁡30∘)=μsmg\mathrm{F}\left(\cos 30^{\circ}+\mu_{\mathrm{s}} \sin 30^{\circ}\right)=\mu_{\mathrm{s}} \mathrm{mg}

F=μsmgcos⁡30∘+μssin⁡30∘=0.25×10×103/2×0.25×1/2\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  F=253/2+0.252=501.73+0.25=501.98=25.2 N\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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