Q69 / 75 · JEE Main 2026 (28 January Shift 1) · Physics
A block of mass 5 kg is moving on an inclined plane which makes an angle of $30^{\circ}$ with the horizontal. Friction coefficient between the block and inclined plane surface is $\frac{\sqrt{3}}{2}$. The force to be applied on the block so that the block will move down without acceleration is $\_\_\_\_$ N.
$\left(g=10 \mathrm{~m} / \mathrm{s}^{2}\right)$.
$\left(g=10 \mathrm{~m} / \mathrm{s}^{2}\right)$.
- A25
- B7.5
- Correct answerC12.5
- D15
Solution
For motion without acceleration, net force = 0.
Component of weight along incline: $mg\sin(30°) = 5 \times 10 \times 0.5 = 25$ N (down).
Normal force: $N = mg\cos(30°) = 50 \times \frac{\sqrt{3}}{2} = 25\sqrt{3}$ N.
Friction force opposing downward motion:
$f = \mu N = \frac{\sqrt{3}}{2} \times 25\sqrt{3} = 37.5$ N (up).
For equilibrium along the incline, applying force F down the incline:
$mg\sin\theta + F = f$, so $25 + F = 37.5$, giving $F = 12.5$ N.
Component of weight along incline: $mg\sin(30°) = 5 \times 10 \times 0.5 = 25$ N (down).
Normal force: $N = mg\cos(30°) = 50 \times \frac{\sqrt{3}}{2} = 25\sqrt{3}$ N.
Friction force opposing downward motion:
$f = \mu N = \frac{\sqrt{3}}{2} \times 25\sqrt{3} = 37.5$ N (up).
For equilibrium along the incline, applying force F down the incline:
$mg\sin\theta + F = f$, so $25 + F = 37.5$, giving $F = 12.5$ N.
Hence, the correct answer is option: 'C'.