Physics Q4 – NEET-UG 2025

Home Physics Laws of MotionQuestion #219

There are two inclined surfaces of equal length (

L

) and same angle of inclination

45^{\circ}

with the horizontal. One of them is rough and the other is perfectly smooth. A given body takes 2 times as much time to slide down on rough surface than on the smooth surface. The coefficient of kinetic friction (

μ_{k}

) between the object and the rough surface is close to

Clue 1 of 3: Understand Motion on an Inclined Plane

The acceleration of a body sliding down an inclined plane depends on the angle of inclination and the presence of friction.

🧠 Full Solution Path

Step 1: Determine Acceleration for Smooth and Rough Surfaces✦ Active

For a smooth inclined plane, the acceleration ( $a_{s}$ ) is given by $a_{s} = g \sin θ$ . Given $θ = 45^{\circ}$ , so $a_{s} = g \sin 45^{\circ} = \frac{g}{\sqrt{2}}$ . For a rough inclined plane, the acceleration ( $a_{r}$ ) is given by $a_{r} = g (\sin θ - μ_{k} \cos θ)$ . Given $θ = 45^{\circ}$ , so $a_{r} = g (\sin 45^{\circ} - μ_{k} \cos 45^{\circ}) = \frac{g}{\sqrt{2}} (1 - μ_{k})$ .

💡 Teacher's Secret Hint

Remember that friction opposes motion, hence the negative sign in the acceleration formula for the rough surface.

Step 2: Apply Kinematic Equation and Relate Times○ Expand

The body starts from rest, so the distance $L$ covered is $L = \frac{1}{2} a t^{2}$ . Let $t_{s}$ be the time taken on the smooth surface and $t_{r}$ be the time taken on the rough surface. We are given $t_{r} = 2 t_{s}$ .

L = \frac{1}{2} a_{s} t_{s}^{2} ⟹ L = \frac{1}{2} (\frac{g}{\sqrt{2}}) t_{s}^{2} (1)

L = \frac{1}{2} a_{r} t_{r}^{2} ⟹ L = \frac{1}{2} (\frac{g}{\sqrt{2}} (1 - μ_{k})) (2 t_{s})^{2} (2)

💡 Teacher's Secret Hint

Ensure you correctly substitute $t_{r} = 2 t_{s}$ into the equation for the rough surface.

Step 3: Solve for the Coefficient of Kinetic Friction (

μ_{k}

)○ Expand

Equating equations (1) and (2) for $L$ :

\frac{1}{2} (\frac{g}{\sqrt{2}}) t_{s}^{2} = \frac{1}{2} (\frac{g}{\sqrt{2}} (1 - μ_{k})) 4 t_{s}^{2}

Cancel out common terms $\frac{1}{2} \frac{g}{\sqrt{2}} t_{s}^{2}$ from both sides:

1 = (1 - μ_{k}) 4

1 = 4 - 4 μ_{k}

4 μ_{k} = 3

μ_{k} = \frac{3}{4} = 0.75

Thus, the coefficient of kinetic friction is $0.75$ .

💡 Teacher's Secret Hint

Double-check your algebraic manipulation, especially when distributing the factor of 4.

✦ STEM Console utilizes AI models to generate step-by-step explanations and math clues. AI can make mistakes.

Physics Question 4 – NEET-UG 2025

Clue 1 of 3: Understand Motion on an Inclined Plane

🧠 Full Solution Path