JEE MAIN - Physics (2022 - 24th June Morning Shift - No. 25)
A ball of mass 100 g is dropped from a height h = 10 cm on a platform fixed at the top of a vertical spring (as shown in figure). The ball stays on the platform and the platform is depressed by a distance $${h \over 2}$$. The spring constant is _____________ Nm$$-$$1.
(Use g = 10 ms$$-$$2)
_24th_June_Morning_Shift_en_25_1.png)
Answer
120
Explanation
$$mg\left( {h + {h \over 2}} \right) = {1 \over 2}k{\left( {{h \over 2}} \right)^2}$$
$$ \Rightarrow 0.1 \times 10\times(0.15) = {1 \over 2}k{(0.05)^2}$$
$$ \Rightarrow k = 120$$ N/m
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