A ball of mass 100 ,g is dropped from a height h = 10, cm on a platform fixed at top of vertical spr

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5.Work, Energy, Power and Collision

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A ball of mass $100 \,g$ is dropped from a height $h =$ $10\, cm$ on a platform fixed at the top of vertical spring (as shown in figure). The ball stays on the platform and the platform is depressed by a distance $\frac{ h }{2}$. The spring constant is.......... $Nm^{-1}$ . (Use $g=10\, ms ^{-2}$ )

A

$122$

B

$129$

C

$127$

D

$120$

(JEE MAIN-2022)

Solution

By energy conservation

$PE = KE$

$mg \left( H +\frac{ H }{2}\right)=\frac{1}{2} kx ^{2}\left( x =\frac{ H }{2}\right)$

$0.100 \times 10 \times \frac{3}{2}(0.10)=\frac{1}{2} k (0.05 \times 0.05)$

$k =\frac{3 \times 0.10}{0.05 \times 0.05}$

$=\frac{3 \times 1000}{25}=120\, N / m$

Standard 11

Physics

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