A block of mass 200 g is kept stationary on a smooth inclined plane by applying a minimum horizontal force F = √xN as shown in figure.
The value of x = _____________.

Solution

Physics sir · 4 steps

Resolve along the incline: the normal force drops out, and F only has to cancel gravity's pull down the slope.

1. Three forces. The block feels its weight mg straight down, the normal force N perpendicular to the incline, and the horizontal push F. The plane is smooth, so there is no friction.
2. Pick the axis. Resolve along the incline, because N has no component in that direction. Gravity pulls down the slope with mg sin 60°. The horizontal F makes 60° with the incline, so it pushes up the slope with F cos 60°.

   F cos 60° = mg sin 60° ⇒ F = mg tan 60°

3. Put in numbers. With m = 0.2 kg and g = 10 m s⁻², F = 0.2 × 10 × √3 = 2√3 N.
4. So. F = 2√3 = √(4 × 3) = √12 N, so x = 12.

The common slip: 3.46. Reporting the force F = 2√3 ≈ 3.46 N gives 3.46, not 12. The question asks for x, where F = √x, so square F to get x.

The idea to keep. On a smooth incline, resolve along the surface so the unknown normal force drops out. A horizontal force holding a block on an incline of angle θ is F = mg tan θ.

Answer from the NTA final answer key. Solution by PrepWiser.