A cubic block of mass m is sliding down on an inclined plane at 60 degrees with an acceleration of g/2, the value of coefficient of kinetic friction is

The correct answer is Option 4. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Physics Q30 – JEE-MAIN 2025

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Home Physics Laws of MotionQuestion #229

A cubic block of mass

m

is sliding down on an inclined plane at

60^{\circ}

with an acceleration of

\frac{g}{2}

, the value of coefficient of kinetic friction is

Clue 1 of 3: Free Body Diagram

Identify all forces acting on the block and resolve them into components parallel and perpendicular to the inclined plane.

🧠 Full Solution Path

Step 1: Identify and Resolve Forces✦ Active

Draw a free-body diagram for the block on the inclined plane. Resolve the gravitational force $m g$ into components parallel ( $m g \sin θ$ ) and perpendicular ( $m g \cos θ$ ) to the plane. The angle of inclination is $θ = 60^{\circ}$ .

Step 2: Apply Newton's Second Law Perpendicular to the Plane○ Expand

Since there is no acceleration perpendicular to the plane, the normal force $N$ balances the perpendicular component of gravity.

N = m g \cos 60^{\circ} = m g (\frac{1}{2}) = \frac{m g}{2}

Step 3: Apply Newton's Second Law Parallel to the Plane○ Expand

The net force along the inclined plane causes the acceleration $a = \frac{g}{2}$ . The forces are the component of gravity down the plane and the kinetic friction force up the plane.

m g \sin 60^{\circ} - f_{k} = m a

m g \sin 60^{\circ} - μ_{k} N = m (\frac{g}{2})

Substitute $N = \frac{m g}{2}$ and $\sin 60^{\circ} = \frac{\sqrt{3}}{2}$ :

m g \frac{\sqrt{3}}{2} - μ_{k} \frac{m g}{2} = m \frac{g}{2}

Divide by $m g / 2$ :

\sqrt{3} - μ_{k} = 1

μ_{k} = \sqrt{3} - 1

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Physics Question 30 – JEE-MAIN 2025

Clue 1 of 3: Free Body Diagram

🧠 Full Solution Path