A body of mass 0.5 kg travels on straight line path with velocity v=(3 x²+4) m / s. The net workdone by the force during its displacement from x=0 to x=2 m is :

Solution

Physics sir · 3 steps

Net work is just the change in kinetic energy, so find the speed at both ends and subtract.

1. Speeds at the ends. At x = 0, v₁ = 3(0)² + 4 = 4 m s⁻¹. At x = 2 m, v₂ = 3(2)² + 4 = 16 m s⁻¹.
2. Work-energy theorem. The net work done by all forces equals the change in kinetic energy. You do not need the force itself.

   W = 1/2m(v₂² - v₁²) = 1/2× 0.5 × (256 - 16)

3. So. W = 0.25 × 240 = 60 J, which is option (b).

The idea to keep. When velocity is given as a function of position, do not hunt for the force. Net work is the change in KE, so you only need the speeds at the start and the end.

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