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QJEEM22Ph776e: JEE Main 2022 Physics - Class 11 JEEM22Ph776e Solution: A 0.5 kg block moving at a speed of 12 ms 1 compresses a spring …

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.

Step-by-Step Verified Solution

12m V2=12kx2+12m(v2)2{1 \over 2}m\,{V^2} = {1 \over 2}k{x^2} + {1 \over 2}m{\left( {{v \over 2}} \right)^2}

⇒38mv2=12kx2\Rightarrow {3 \over 8}m{v^2} = {1 \over 2}k{x^2}

⇒k=34×12×1449×100\Rightarrow k = {3 \over 4} \times {1 \over 2} \times {{144} \over 9} \times 100

=600= 600

⇒600\Rightarrow 600

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