A block of mass 1 kg , moving along x with speed vᵢ=10 m / s enters a rough region ranging from x=0.1 m to x=1.9 m. The retarding force acting on the block in this range is Fᵣ=-kr N, with k =10 N / m. Then the final speed of the block as it crosses rough region is.

Solution

Physics sir · 3 steps

The force grows with x, so integrate it for the work, then let the work-energy theorem find the speed.

1. Variable force. The retarding force is F = -kx, where x is the block's position. Its size changes along the path, so the work done is an integral, not force times distance.

   W = -∫_(0.1)^(1.9) kx dx = -k/2(1.9² - 0.1²) = -5 × 3.6 = -18 J

2. Work-energy theorem. The initial kinetic energy is 1/2 × 1 × 10² = 50 J. The rough region removes 18 J, so the final kinetic energy is 50 - 18 = 32 J.
3. So. 1/2 × 1 × v_f² = 32 gives v_f² = 64, so v_f = 8 m/s, which is option (c).

The idea to keep. When a force depends on position, find its work with ∫ F dx between the exact limits. Then W = Δ K gives the speed without needing the acceleration.

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