A block of mass 1 kg is pushed up a surface inclined to horizontal at an angle of 60° by a force of 10 N parallel to the inclined surface as shown in figure. When the block is pushed up by 10 m along inclined surface, the work done against frictional force is :
[g=10 m / s²]

Solution

Physics sir · 3 steps

Friction comes from the normal force alone, so find N = mg cos 60° first.

1. Normal force. The 10 N push is parallel to the surface, so it adds nothing to the normal force. Only the component of weight perpendicular to the incline presses the block in: N = mg cos 60° = 1 × 10 × 1/2 = 5 N.
2. Friction force. With μ = 0.1, the friction force is f = μ N = 0.1 × 5 = 0.5 N, acting down the incline. The block does move up, because 10 N is more than mg sin 60° + f ≈ 8.66 + 0.5 = 9.16 N.

   W_(against friction) = f × d = μ mg cos θ × d

3. So. W = 0.5 × 10 = 5 J, which is option (b).

The idea to keep. On an incline, friction is μ mg cos θ whenever the applied force is along the surface. The sin θ part of the weight pulls the block along the slope. It never sets the friction.

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