A ball of mass $m$ is dropped from height $h$ on a light platform fixed at the top of vertical spring as shown in the figure. The maximum compression of spring is $x$. What is the spring constant ?

2 m g h + x

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\frac{2 m g (h + x)}{h^{2}}

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\frac{2 m g (h + x)}{x^{2}}

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\frac{2 m g h}{x^{2}}

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Solution

The correct option is C $\frac{2 m g (h + x)}{x^{2}}$
Given:
$mass of ball = m$
$plateform depressed by = x$
$ball is dropped from h as shown in figure$
By work energy theorem
$(W . D)_{g} + (W . D)_{spring} = Δ K E$
$= K_{F} - K_{i}$
$m g (h + x) - \frac{1}{2} k x^{2} = 0 - 0$
$k = \frac{2 m g (h + x)}{x^{2}}$
Hence, option $(C)$ is correct.

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A ball of mass \(m\) is dropped from height \(h\) on a light platform fixed at the top of vertical spring as shown in the figure. The maximum compression of spring is \(x\). What is the spring constant ?

The correct option is C 2mg(h+x)x2Given: mass of ball=m plateform depressed by=x ball is dropped from h as shown in figure By work energy theorem (W.D)g+(W.D)spring=ΔKE =KF−Ki mg(h+x)−12kx2=0−0 k=2mg(h+x)x2 Hence, option (C) is correct.