A 0.5 kg block moving at a speed of 12 ms⁻¹ compresses a spring through a distance 30 cm when its speed is halved. The spring constant of the spring will be _______________ Nm⁻¹.

Solution

Physics sir · 4 steps

The block does not stop, so only the kinetic energy it loses goes into the spring.

1. Energy lost by block. The speed falls from 12 m s⁻¹ to 6 m s⁻¹. The loss in kinetic energy is 1/2 × 0.5 × (12² - 6²) = 0.25 × 108 = 27 J.
2. Energy gained by spring. All of this lost energy is stored in the spring. The compression is x = 30 cm = 0.3 m, so x² = 0.09 m².

   1/2m(u² - v²) = 1/2kx²

3. Solve for k. 27 = 1/2 × k × 0.09, so k = 54/0.09 = 600 N m⁻¹.
4. So. The spring constant is 600 N m⁻¹.

The common slip: 800. Using the full initial kinetic energy of 36 J, as if the block stopped, gives 800 instead of 600.

The idea to keep. When a moving body only slows down against a spring, set the change in kinetic energy equal to 1/2kx². Always convert the compression to metres first.

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