A particle of mass m = 1 kg is dropped from a height h = 40 cm on a light horizontal platform fixed to one end of an elastic spring, the other being fixed to a base, as shown in the diagram. The parti

Question

A particle of mass m = 1 kg is dropped from a height h = 40 cm on a light horizontal platform fixed to one end of an elastic spring, the other being fixed to a base, as shown in the diagram. The particle collides with the platform and sticks to it. As a result, the spring is compressed by an amount x = 10 cm . What is the force constant of the spring? (Take g = 10 m s - 2 )

TestHub · Accepted Answer

Since the platform is depressed by an amount x, the total work done on the spring is $m g (h + x)$ . This work is stored in the spring in the form of potential energy $\frac{1}{2} k x^{2}$ .

Equating the two, we have

$\frac{1}{2} k x^{2} = m g (h + x)$ or $k = \frac{2 mg (h + x)}{x^{2}}$

Given, $h = 0.4 m$ , $x = 0.1 m$ , $m = 1 kg$ and $g = 10 m s^{- 2}$ . Substituting these values, we get
$k = 1000 N m^{- 1}$ .

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