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 $\frac{h}{2}$ . The spring constant is ${Nm}^{- 1}$ .
(Use $g = 10 {ms}^{- 2}$ )

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Solution

From energy conservation principle: $m g (h + \frac{h}{2}) = \frac{1}{2} k {(\frac{h}{2})}^{2}$
$\Rightarrow 0.1 \times 10 \times (0.15) = \frac{1}{2} k (0.05)^{2}$
$\Rightarrow k = 120 N/m$

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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 a vertical spring (as shown in figure). The ball stays on the platform and the platform is depressed by a distance h2. The spring constant is Nm−1. (Use g=10 ms−2)

From energy conservation principle: mg(h+h2)=12k(h2)2 ⇒0.1×10×(0.15)=12k(0.05)2 ⇒k=120 N/m

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 $\frac{h}{2}$ . The spring constant is ${Nm}^{- 1}$ .
(Use $g = 10 {ms}^{- 2}$ )

From energy conservation principle: $m g (h + \frac{h}{2}) = \frac{1}{2} k {(\frac{h}{2})}^{2}$
$\Rightarrow 0.1 \times 10 \times (0.15) = \frac{1}{2} k (0.05)^{2}$
$\Rightarrow k = 120 N/m$