Work, Power and Energy Part 1

Given, the length of the shock absorber, I = 1.5 m The total mass of the system, M = 40000 kg The speed of the wagon, v = 72 km/h When brakes are applied, the final velocity, $v_f$ = 0 The compressed spring of the shock absorber, x = 1mApplying the work-energy theorem, Work done by the system = Change in kinetic energy W = Δ KE $W_{\text{friction}}+W_{\text{spring}}=\frac{1}{2} m v_{f}^{2}+\frac{1}{2} m v_{i}^{2}$ $-\frac{90}{100}\left(\frac{1}{2} m v^{2}\right)+W_{\text{spring}}=0-\frac{1}{2} m v_{i}^{2}$ (∵ 90% energy lost due to friction) $W_{\text{spring}}=-\frac{10}{100} \times \frac{1}{2} m v^{2}$ $-\frac{1}{2} k x^{2}=\frac{1}{20} m v^{2}$ $k=\frac{m v^{2}}{10 \times x^{2}}$ Substituting the values in the above equation, we get $k=\frac{40000 \times \left(72 \times \frac{5}{18}\right)^{2}}{10(1)^{2}}$ = 16 × $10^{5}$ N / m Comparing the spring constant, k = x × $10^{5}$ The value of the x = 16.