A block of mass 5 kg is placed on a rough inclined surface as shown in the figure. If F₁⃗ is the force required to just move the block up the inclined plane and F₂⃗ is the force required to just prevent the block from sliding down, then the value of |F₁⃗|-|F₂⃗| is : [Use g=10 m / s²]

Solution

Physics sir · 4 steps

Going up, friction fights you; holding back, friction helps you, so the gap is just twice the friction.

1. Forces along incline. Take F₁ and F₂ along the incline. Gravity pulls the block down the slope with mg sin 30° = 5 × 10 × 1/2 = 25 N. The normal force is N = mg cos 30° = 25√3 N, so limiting friction is f = μ N = 0.1 × 25√3 = 2.5√3 N.
2. Just moving up. The block is about to slide up, so friction acts down the slope along with gravity. Then F₁ = mg sin θ + f = 25 + 2.5√3 N.
3. Just held back. Since tan 30° = 0.577 is more than μ = 0.1, the block would slide down on its own. When it is about to slide down, friction acts up the slope and helps F₂. Then F₂ = mg sin θ - f = 25 - 2.5√3 N.

   |F₁⃗| - |F₂⃗| = 2μ mg cos θ = 2 × 2.5√3 = 5√3 N

4. So. The mg sin θ terms cancel and the difference is 5√3 N, option (a).

The idea to keep. Friction always opposes the motion that is about to happen. Pushing up needs mg sin θ + f, holding back needs mg sin θ - f, so their difference is 2f.

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