JEE Main · 2021 · 01 Sep 2021 · Shift 2 · Work, Power & Energy

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 1 0^{5} N / m

Answer: 16

Solution

The wagon's kinetic energy is

\frac{1}{2} m v^{2} = \frac{1}{2} \times 40000 \times 2 0^{2} = 8 \times 1 0^{6} J

v = 72 km/h = 20 m/s

. With

90%

lost to friction, the spring stores

0.1 \times 8 \times 1 0^{6} = 8 \times 1 0^{5} J

. Setting

\frac{1}{2} k x^{2} = 8 \times 1 0^{5}

with the compression

x = 1 m

gives the spring constant

16 \times 1 0^{5} N/m

, so the required value is

16

JEE Main · 2021 · 01 Sep 2021 · Shift 2 · Integer answer

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JEE Main · 2021 · 01 Sep 2021 · Shift 2 · Work, Power & Energy

Solution

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