An object of mass 'm' initially at rest on a smooth horizontal plane starts moving under the action of force F = 2N. In the process of its linear motion, the angle θ (as shown in figure) between the direction of force and horizontal varies as θ=kx, where k is a constant and x is the distance covered by the object from its initial position. The expression of kinetic energy of the object will be E = n/k sin θ. The value of n is ___________.

Solution

Physics sir · 4 steps

The angle keeps changing, so add up F cos θ dx bit by bit; that integral is the kinetic energy.

1. Only horizontal part works. The object moves along the horizontal plane. The vertical component F sin θ is perpendicular to the motion and does no work. Only F cos θ along the motion does work.
2. Variable force. Since θ = kx changes with position, F cos θ is not constant. Find the work over a tiny displacement dx, which is dW = F cos(kx) dx, and integrate from 0 to x.

   W = ∫₀ˣ 2 cos(kx) dx = 2/k sin(kx) = 2/k sin θ

3. Work-energy theorem. The plane is smooth and the object starts from rest, so all this work becomes kinetic energy: E = 2/k sin θ.
4. So. Comparing with E = n/k sin θ gives n = 2.

The idea to keep. When a force changes direction with position, never use Fx cos θ directly. Write dW = F cos θ dx and integrate, then use the work-energy theorem.

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