Physics Q28 – JEE-MAIN 2026

Home Physics Laws of MotionQuestion #212

The time taken by a block of mass

m

to slide down from the highest point to the lowest point on a rough inclined plane is

50 %

more compared to the time taken by the same block on identical inclined smooth plane. Both inclined planes are at

45^{\circ}

with the horizontal. The coefficient of kinetic friction between the rough inclined surface and block is _______. .

Clue 1 of 3: Kinematics and Forces

Analyze the forces acting on the block on both smooth and rough inclined planes to determine the acceleration in each case.

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Step 1: Determine accelerations for smooth and rough planes✦ Active

Let $L$ be the length of the inclined plane. For a smooth inclined plane, the net force along the plane is $m g \sin θ$ . The acceleration is $a_{s} = g \sin θ$ . For a rough inclined plane, the kinetic friction force is $f_{k} = μ_{k} N = μ_{k} m g \cos θ$ . The net force is $m g \sin θ - μ_{k} m g \cos θ$ . The acceleration is $a_{r} = g (\sin θ - μ_{k} \cos θ)$ .

Step 2: Relate time and acceleration using kinematics○ Expand

Since the block starts from rest, the distance $L$ covered in time $t$ is given by $L = \frac{1}{2} a t^{2}$ . This implies $t = \sqrt{\frac{2 L}{a}}$ . Therefore, the time taken for the smooth plane is $t_{s} = \sqrt{\frac{2 L}{g \sin θ}}$ and for the rough plane is $t_{r} = \sqrt{\frac{2 L}{g (\sin θ - μ_{k} \cos θ)}}$ .

Step 3: Use the given time relationship to find

μ_{k}

○ Expand

We are given that $t_{r} = 50 %$ more than $t_{s}$ , so $t_{r} = t_{s} + 0.5 t_{s} = 1.5 t_{s} = \frac{3}{2} t_{s}$ . Squaring both sides, $t_{r}^{2} = \frac{9}{4} t_{s}^{2}$ . Substituting the expressions for $t_{s}^{2}$ and $t_{r}^{2}$ :

\frac{2 L}{g (\sin θ - μ_{k} \cos θ)} = \frac{9}{4} (\frac{2 L}{g \sin θ})

\frac{1}{\sin θ - μ_{k} \cos θ} = \frac{9}{4 \sin θ}

4 \sin θ = 9 (\sin θ - μ_{k} \cos θ)

4 \sin θ = 9 \sin θ - 9 μ_{k} \cos θ

9 μ_{k} \cos θ = 5 \sin θ

μ_{k} = \frac{5 \sin θ}{9 \cos θ} = \frac{5}{9} \tan θ

Given $θ = 45^{\circ}$ , so $\tan 45^{\circ} = 1$ . Therefore, $μ_{k} = \frac{5}{9} \times 1 = \frac{5}{9}$ .

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Physics Question 28 – JEE-MAIN 2026

Clue 1 of 3: Kinematics and Forces

🧠 Full Solution Path