Question

An engine is attached to a wagon through a shock absorber of length 1.5m. the system with a total mass of 50,000 kg is moving with a speed of 36 km/hr 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, calculate the spring constant.

Answer 1

m = 50,000 kg

v = 36×\(\frac{5}{18}\)m/s = 10 m/s

KE = \(\frac{1}{2}\)mv² = \(\frac{1}{2}\) × 5 × 10⁴×10² J

= 2.5 × 10⁶ J

90% of KE of wagon lost due to friction and only 10% of this is stored in the spring.

\(\frac{1}{2}\)kx² = 2.5×10⁴ = 10% of 2.5 × 10⁶ J

Here, x = 1 m

\(So,k=\frac{\frac{2\times 2.5 \times 10^4}{(1)^2}N}{m}\)

k = 5.0 × 10⁴ N/m