JEE MAIN - Physics (2022 - 25th June Morning Shift - No. 20)
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
Explanation
$${1 \over 2}m\,{V^2} = {1 \over 2}k{x^2} + {1 \over 2}m{\left( {{v \over 2}} \right)^2}$$
$$ \Rightarrow {3 \over 8}m{v^2} = {1 \over 2}k{x^2}$$
$$ \Rightarrow k = {3 \over 4} \times {1 \over 2} \times {{144} \over 9} \times 100$$
$$ = 600$$
$$ \Rightarrow 600$$
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