Consider a block kept on an inclined plane (inclined at $45^{\circ}$) as shown in the figure. If the force required to just push it up the incline is $2$ times the force required to just prevent it from sliding down, the coefficient of friction between the block and inclined plane $(\mu )$ is equal to : ![JEE Main 2023 Physics, Laws of Motion — question figure](https://prepforbharat.s3.ap-south-1.amazonaws.com/exam/615f0e999476412f48314daf/Physics/images/Laws_of_Motion/648b5a6f417cc3fb48d680be/question_1__q_648b5a6f417cc3fb48d680be__cdn-question-pool.getmarks.app__1973720d-453b-478c-86d0-f14845e7a8f7-9894326__15a9f24339_final_ppt_sync.png).

Consider a block kept on an… - JEE Main 2023 Physics PYQ

JEE Main 2023 — Physics Mechanics

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Consider a block kept on an inclined plane (inclined at $45^{\circ}$ ) as shown in the figure. If the force required to just push it up the incline is $2$ times the force required to just prevent it from sliding down, the coefficient of friction between the block and inclined plane $(\mu )$ is equal to :

JEE Main 2023 Physics, Laws of Motion — question figure .

Official previous-year question

Held on 25 Jan 2023 · Verified 6 Jul 2026.

Options

A
$0.33$
B
$0.60$
C
$0.25$
D
$0.50$

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Solution

JEE Main 2023 Physics, Laws of Motion — solution figure 1

The forces acting on block is shown in above figure.

Here, normal force, $N=mg\mathrm{cos}45^{\circ}$ and frictional force, $f=\mu N$ .

Balancing the force along incline, we have

${F}_{1}=mg\mathrm{sin}45^{\circ}+f=mg\mathrm{sin}45^{\circ}+\mu N$

$\Rightarrow {F}_{1}=\frac{mg}{\sqrt{2}}+\mu mg\mathrm{cos}45^{\circ}$

$\Rightarrow {F}_{1}=\frac{mg}{\sqrt{2}}(1+\mu )$

Now, the forces acting on block to just push it up on inclined plane is shown below.

JEE Main 2023 Physics, Laws of Motion — solution figure 2

Balancing the force along incline, we have

${F}_{2}=mg\mathrm{sin}45^{\circ}-f=mg\mathrm{sin}45^{\circ}-\mu N$

$=\frac{mg}{\sqrt{2}}(1-\mu )$

Given, ${F}_{1}=2{F}_{2}$

$\Rightarrow \frac{mg}{\sqrt{2}}(1+\mu )=2\frac{mg}{\sqrt{2}}(1-\mu )$

$\Rightarrow 1+\mu =2-2\mu$

$\Rightarrow \mu =\frac{1}{3}=0.33$

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Solution

JEE Main 2023 Physics, Laws of Motion — solution figure 1

The forces acting on block is shown in above figure.

Here, normal force, $N=mg\mathrm{cos}45^{\circ}$ and frictional force, $f=\mu N$ .

Balancing the force along incline, we have

${F}_{1}=mg\mathrm{sin}45^{\circ}+f=mg\mathrm{sin}45^{\circ}+\mu N$

$\Rightarrow {F}_{1}=\frac{mg}{\sqrt{2}}+\mu mg\mathrm{cos}45^{\circ}$

$\Rightarrow {F}_{1}=\frac{mg}{\sqrt{2}}(1+\mu )$

Now, the forces acting on block to just push it up on inclined plane is shown below.

JEE Main 2023 Physics, Laws of Motion — solution figure 2

Balancing the force along incline, we have

${F}_{2}=mg\mathrm{sin}45^{\circ}-f=mg\mathrm{sin}45^{\circ}-\mu N$

$=\frac{mg}{\sqrt{2}}(1-\mu )$

Given, ${F}_{1}=2{F}_{2}$

$\Rightarrow \frac{mg}{\sqrt{2}}(1+\mu )=2\frac{mg}{\sqrt{2}}(1-\mu )$

$\Rightarrow 1+\mu =2-2\mu$

$\Rightarrow \mu =\frac{1}{3}=0.33$

Options

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Solution

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JEE Main Mechanics in other years

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Options

Did you get this right?

Solution

Your note

JEE Main Mechanics in other years

More JEE Main Mechanics questions

Other Physics topics for JEE Main

Electromagnetism

Modern Physics

Optics

Thermodynamics

Waves & Oscillations

Keep practicing Mechanics