Ncert Exercises & Solutions

Q. 41 An engine is attached to a wagon through a shock absorber of length 1.5 m. The system with a total mass of 50,000 kg is moving with a speed of 36 km/h 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 the energy of the wagon is lost due to friction, calculate the spring constant.

Hint: Loss in kinetic energy = work done against friction + energy of shock absorber.

Step 1: Find the kinetic energy of the system.

Given, the mass of the system (m) = 50,000 kg
Speed of the system (v) = 36 km/h
$= \frac{36 \times 1000}{60 \times 60} = 10 m / s$
Compression of the spring (x) = 10m
$\begin{array}{l} KE of the system = \frac{1}{2} {mv}^{2} \\ = \frac{1}{2} \times 50000 \times (10)^{2} \\ = 25000 \times 100 J = 2 .5 \times 10^{6} J \end{array}$

Step 2: Find spring constant.
Since, 90% of KE of the system is lost due to friction, therefore. energy to shock absorber is given by

$\begin{array}{l} ΔE = \frac{1}{2} {kx}^{2} = 10 % of total KE of the system \\ = \frac{10}{100} \times 2 .5 \times 10^{6} J or k = \frac{2 \times 2 .5 \times 10^{6}}{10 \times (1)^{2}} \\ = 5 .0 \times 10^{5} N / m \end{array}$

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