JEE MAIN - Physics (2021 - 1st September Evening Shift - No. 28)
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 ____________ $$\times$$ 105 N/m.
Answer
16
Explanation
Given, the length of the shock absorber, l = 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, vf = 0
The compressed spring of the shock absorber, x = 1 m
Applying the work-energy theorem,
Work done by the system = Change in kinetic energy
$$W = \Delta KE$$
$${W_{friction}} + {W_{spring}} = {1 \over 2}mv_f^2 + {1 \over 2}mv_i^2$$
$$ - {{90} \over {100}}\left( {{1 \over 2}m{v^2}} \right) + {W_{spring}} = 0 - {1 \over 2}mv_i^2$$ ($$\because$$ 90% energy lost due to friction)
$${W_{spring}} = - {{10} \over {100}} \times {1 \over 2}m{v^2}$$
$$ - {1 \over 2}k{x^2} = {1 \over {20}}m{v^2}$$
$$k = {{m{v^2}} \over {10 \times {x^2}}}$$
Substituting the values in the above equation, we get
$$k = {{40000 \times {{\left( {72 \times {5 \over {18}}} \right)}^2}} \over {10{{(1)}^2}}}$$
= 16 $$\times$$ 105 N/m
Comparing the spring constant, k = x $$\times$$ 105
The value of the x = 16.
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, vf = 0
The compressed spring of the shock absorber, x = 1 m
Applying the work-energy theorem,
Work done by the system = Change in kinetic energy
$$W = \Delta KE$$
$${W_{friction}} + {W_{spring}} = {1 \over 2}mv_f^2 + {1 \over 2}mv_i^2$$
$$ - {{90} \over {100}}\left( {{1 \over 2}m{v^2}} \right) + {W_{spring}} = 0 - {1 \over 2}mv_i^2$$ ($$\because$$ 90% energy lost due to friction)
$${W_{spring}} = - {{10} \over {100}} \times {1 \over 2}m{v^2}$$
$$ - {1 \over 2}k{x^2} = {1 \over {20}}m{v^2}$$
$$k = {{m{v^2}} \over {10 \times {x^2}}}$$
Substituting the values in the above equation, we get
$$k = {{40000 \times {{\left( {72 \times {5 \over {18}}} \right)}^2}} \over {10{{(1)}^2}}}$$
= 16 $$\times$$ 105 N/m
Comparing the spring constant, k = x $$\times$$ 105
The value of the x = 16.
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