JEE Main · 2022 · 25 Jun 2022 · Shift 1 · Work, Power & Energy

A 0.5 kg block moving at a speed of

12 ms^{- 1}

compresses a spring through a distance 30 cm when its speed is halved. The spring constant of the spring will be ______

Nm^{- 1}

Answer: 600

Solution

Using conservation of energy:

K_{i} + U_{i} = K_{f} + U_{f}

Initially, the spring is uncompressed (

U_{i} = 0

) and the block has kinetic energy

\frac{1}{2} m (12)^{2}

.

Finally, the spring is compressed by 0.3 m and the block's speed is halved to 6 m/s:

\frac{1}{2} (0.5) (12)^{2} = \frac{1}{2} k (0.3)^{2} + \frac{1}{2} (0.5) (6)^{2}

0.5 (144 - 36) = k (0.09)

0.5 (108) = 0.09 k

k = \frac{54}{0.09} = 600 N/m

JEE Main · 2022 · 25 Jun 2022 · Shift 1 · Integer answer

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