A particle of mass 500 gm is moving in a straight line with velocity v = b x 5/2 . The work done by the net force during its displacement from x = 0 to x = 4 m is : (Take b = 0.25 m -3/2 s⁻¹).

Solution

Physics sir · 4 steps

Do not hunt for the force. Net work is just the change in kinetic energy, so find the speeds at both ends.

1. Use work-energy theorem. The work done by the net force equals the change in kinetic energy. So you only need the speed at x = 0 and at x = 4 m.

   Wₙₑₜ = 1/2mv_f² - 1/2mvᵢ²

2. Find both speeds. At x = 0, vᵢ = b × 0 = 0. At x = 4 m, 4^(5/2) = (√4)⁵ = 2⁵ = 32, so v_f = 0.25 × 32 = 8 m s⁻¹.
3. Put in numbers. With m = 500 g = 0.5 kg, Wₙₑₜ = 1/2 × 0.5 × 8² - 0 = 0.25 × 64 = 16 J.
4. So. The work done by the net force is 16 J, option (d).

The idea to keep. When you know velocity as a function of position, net work is Δ KE between the two points. No need to find the force or integrate.

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