An object of mass 'm' is being moved with a constant velocity under the action of an applied force of 2N along a frictionless surface with following surface profile.
The correct applied force vs distance graph will be :

Solution

Physics sir · 4 steps

Constant velocity means zero net force, so the push only has to cancel mg sin θ on each slope.

1. Net force is zero. The velocity is constant, so the acceleration is zero and the forces along the slope must add to zero. The surface is frictionless, so along the slope only gravity's component mg sin θ and the applied force F act.

   F = mg sin θ = 2 N (directed up the slope)

2. Going up. On the left slope, mg sin θ pulls the block back down. F must point up the slope, which is the direction of motion, so F = +2 N.
3. Coming down. Both slopes make the same angle θ, so the magnitude stays 2 N. On the right slope, mg sin θ pulls the block forward, down the slope. F must hold it back, pointing up the slope against the motion, so F = -2 N.
4. So. F is +2 N over the first half of D and -2 N over the second half: two flat steps, which is graph (c).

The idea to keep. On a frictionless incline at constant velocity, the push equals mg sin θ, whatever the height. Its sign flips when gravity's component switches from opposing the motion to helping it.

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