A particle is projected with velocity v₀ along x-axis. A damping force is acting on the particle which is proportional to the square of the distance from the origin i.e. ma = - αx². The distance at which the particle stops :

Solution

Physics sir · 4 steps

The force depends on x, not t, so write a = v dv/dx and integrate from start to stop.

1. Choose the right form. The force is given as a function of position, so write the acceleration as a = vdv/dx. Then mvdv/dx = -α x², and the variables v and x separate.
2. Set the limits. At x = 0 the speed is v₀. At the stopping point x = d the speed is 0.

   m∫_(v₀)⁰ v dv = -α∫₀ᵈ x² dx ⇒ -1/2mv₀² = -(α d³)/3

3. Solve for d. This gives d³ = 3mv₀²/2α. It also matches the work-energy theorem: the work done by the force, (α d³)/3, removes all the kinetic energy 1/2mv₀².
4. So. d = (3mv₀²/2α)^(1/3), which is option (a).

The idea to keep. When force or acceleration is given as a function of position, use a = vdv/dx. It is the work-energy theorem in disguise: the integral of force over distance equals the change in kinetic energy.

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