Question

An engine is attached to a wagon through a shock absorber of length $1.5 \, m$ . The entire system with a total mass of $50000 \, kg$ was moving with a speed of $36 \, kmh^{- 1}$ , when the brakes are applied to bring it to rest. In this process, 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, the spring constant is

• A. $5.0 \times 10^{5} \, \, \text{N/m}$
• B. $4.0 \times 10^{5} \, \, \text{N/m}$
• C. $1.0 \times 10^{5} \, \, \text{N/m}$
• D. $2.0 \times 10^{5} \, \, \text{N/m}$

Answer 1

Answer: (A)

Mass of the system $\left(\right. m \left.\right) = 50000 \, \, \text{kg}$
Speed of the system $\left(\right. v \left.\right) = 36 \, \, \text{km/h}$
$= 3 6 \times \frac{5}{1 8} \text{m/s} \left(\because 1 \text{km/h} = \frac{5}{1 8} \text{m/s}\right)$
$= 10 \, \, \text{m/s}$
Compression of the spring $\left(\right. x \left.\right) = 1.0 \, \, \text{m}$
$\text{KE of the system} = \frac{1}{2} \text{mv}^{2}$
$= \frac{1}{2} \left(\times 5 0 0 0 0 \times \left(\right. 1 0 \left.\right)^{2}$
$= 25000 \times 100 \, \, \text{J}$
$= 2.5 \times 10^{6} \, \, \text{J}$
Since, 90% KE of the system is lost due to friction, therefore kinetic energy transferred to shock absorber,
$= \, 10 \% , \, \, K E_{t o t a l} = \frac{10}{100} \times 2.5 \times 10^{6} \text{J} = 2.5 \times 10^{5} \text{J}$
This K.E. is converted in to spring energy
$2.5 \times 10^{5} = \frac{1}{2} k x^{2}$
$2.5 \times 10^{5} = \frac{1}{2} \times k \times 1^{2}$
$k = 5.0 \times 10^{5} \, \, \text{N/m}$