JEE MAIN - Physics (2022 - 28th July Morning Shift - No. 23)
A block of mass '$$\mathrm{m}$$' (as shown in figure) moving with kinetic energy E compresses a spring through a distance $$25 \mathrm{~cm}$$ when, its speed is halved. The value of spring constant of used spring will be $$\mathrm{nE} \,\,\mathrm{Nm}^{-1}$$ for $$\mathrm{n}=$$ _________.
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Answer
24
Explanation
$$\Delta KE = {W_{all}}$$
So $${E \over 4} - E = - {1 \over 2}K \times {(0.25)^2}$$
$${{3E} \over 4} = {1 \over 2}K \times {1 \over {16}}$$
$$ = K = 24E$$
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