Physics Q86 – AP-EAMCET 2025

Home Physics Motion in a Straight LineQuestion #319

If the time taken for a block to slide down a smooth inclined plane of angle of inclination

30^{\circ}

4 s

, then the time taken by the block to slide down a rough inclined plane of same length and same angle of inclination is (The coefficient of kinetic friction between the inclined plane and the block

= 0.28 \sqrt{3}

)

Clue 1 of 3: Kinematics on an Inclined Plane

The motion of a block sliding down an inclined plane is governed by constant acceleration, which depends on the angle of inclination and the coefficient of friction.

🧠 Full Solution Path

Step 1: Determine the acceleration for a block sliding down an inclined plane✦ Active

The forces acting on a block sliding down an inclined plane are gravity ( $m g \sin θ$ down the plane) and friction ( $f_{k} = μ_{k} N = μ_{k} m g \cos θ$ up the plane). Using Newton's second law, $m a = m g \sin θ - μ_{k} m g \cos θ$ . The acceleration is $a = g (\sin θ - μ_{k} \cos θ)$ . For motion starting from rest, the distance $L$ covered in time $t$ is given by $L = \frac{1}{2} a t^{2}$ . Therefore, the time taken is $t = \sqrt{\frac{2 L}{a}}$ .

Step 2: Calculate the acceleration and derive a relationship for the smooth inclined plane○ Expand

For a smooth inclined plane, the coefficient of kinetic friction $μ_{k} = 0$ . The angle of inclination is $θ = 30^{\circ}$ . The acceleration for the smooth plane ( $a_{1}$ ) is:

a_{1} = g (\sin 30^{\circ} - 0 \cdot \cos 30^{\circ}) = g \sin 30^{\circ} = g \times \frac{1}{2} = \frac{g}{2}

The time taken for the smooth plane ( $t_{1}$ ) is given as $4 s$ . Using $t_{1} = \sqrt{\frac{2 L}{a_{1}}}$ :

4 = \sqrt{\frac{2 L}{g / 2}} = \sqrt{\frac{4 L}{g}}

Squaring both sides:

16 = \frac{4 L}{g} ⟹ \frac{L}{g} = \frac{16}{4} = 4

This value of $\frac{L}{g}$ will be used for the rough plane as the length and angle are the same.

💡 Teacher's Secret Hint

Remember that for a smooth surface, friction is negligible, simplifying the acceleration calculation.

Step 3: Calculate the acceleration for the rough inclined plane○ Expand

For the rough inclined plane, the angle is $θ = 30^{\circ}$ and the coefficient of kinetic friction is $μ_{k} = 0.28 \sqrt{3}$ . The acceleration for the rough plane ( $a_{2}$ ) is:

a_{2} = g (\sin 30^{\circ} - μ_{k} \cos 30^{\circ})

Substitute the values:

a_{2} = g (\frac{1}{2} - 0.28 \sqrt{3} \times \frac{\sqrt{3}}{2})

a_{2} = g (\frac{1}{2} - 0.28 \times \frac{3}{2})

a_{2} = g (0.5 - 0.28 \times 1.5)

a_{2} = g (0.5 - 0.42)

a_{2} = 0.08 g

💡 Teacher's Secret Hint

Be careful with the calculation involving $\sqrt{3} \times \sqrt{3} = 3$ . Ensure correct decimal multiplication.

Step 4: Calculate the time taken for the block to slide down the rough inclined plane○ Expand

Using the formula $t_{2} = \sqrt{\frac{2 L}{a_{2}}}$ and the value of $a_{2} = 0.08 g$ :

t_{2} = \sqrt{\frac{2 L}{0.08 g}}

Rearrange to use the $\frac{L}{g}$ ratio:

t_{2} = \sqrt{\frac{2}{0.08} \times \frac{L}{g}}

Substitute $\frac{L}{g} = 4$ :

t_{2} = \sqrt{\frac{2}{0.08} \times 4}

t_{2} = \sqrt{\frac{8}{0.08}}

t_{2} = \sqrt{100}

t_{2} = 10 s

The time taken for the block to slide down the rough inclined plane is $10 s$ .

💡 Teacher's Secret Hint

Double-check your arithmetic, especially when dividing by decimals. A common mistake is misplacing the decimal point.

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Physics Question 86 – AP-EAMCET 2025

Clue 1 of 3: Kinematics on an Inclined Plane

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