An engine is attached to a wagon through a shock absorber of length 1.5 m. The system with a total mass of 40,000 kg is moving with a speed of 72 kmh⁻¹ when the brakes are applied to bring it to rest. In the process of the system being brought to rest, the spring of the shock absorber gets compressed by 1.0 m. If 90% of energy of the wagon is lost due to friction, the spring constant is ____________ × 10⁵ N/m.

Solution

Physics sir · 4 steps

Friction takes 90% of the kinetic energy, so only the remaining 10% goes into squeezing the spring.

1. Convert the speed. 72 km h⁻¹ = 72 × 5/18 = 20 m s⁻¹.
2. Initial kinetic energy. K = 1/2mv² = 1/2 × 40000 × 20² = 8 × 10⁶ J.
3. Energy in spring. Friction removes 90%, so the spring stores 10% of K, which is 8 × 10⁵ J. It is compressed by x = 1.0 m. The 1.5 m length of the shock absorber is not needed.

   1/2kx² = 0.1 × 1/2mv² ⇒ k = (2 × 8 × 10⁵)/(1.0)² = 16 × 10⁵ N m⁻¹

4. So. The spring constant is 16 × 10⁵ N m⁻¹, so the blank is 16.

The common slip: 144. Giving the spring the 90% that friction takes, instead of the 10% that is left, gives 144, not 16.

The idea to keep. When some energy is lost, ask which share is left for the useful job. Here the spring gets only what friction did not take. Also convert km/h to m/s before you square the speed.

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