Question

PLEASE ANSWER THIS
A body of mass m = 1 kg is dropped from a height h = 40 cm on a horizontal platform fixed to one end of an elastic spring, the other being fixed to a base, as shown in Fig. As a result the spring is compressed by an amount x = 10 cm. What is the force constant of the spring. Take g = 10 ms-2.

Answer 1

Mass of the body, $\sf{m=1\ kg.}$

Height from which the body is dropped, $\sf{h=40\ cm=0.4\ m.}$

Compression in the spring, $\sf{x=10\ cm=0.1\ m.}$

Let the force constant of the spring be $\sf{k.}$

As the spring gets compressed, it attains potential energy due to compression, which is provided by the body dropped over it, according to law of conservation of energy.

The potential energy of the body dropped from the height is,

$\longrightarrow\sf{U_b=mgh}$

$\longrightarrow\sf{U_b=1\times10\times0.4}$

$\longrightarrow\sf{U_b=4\ J}$

This same potential energy is stored in the spring as its potential energy due to compression. Therefore,

$\longrightarrow\sf{\dfrac{1}{2}\,kx^2=4}$

$\longrightarrow\sf{k\times0.1^2=8}$

$\longrightarrow\underline{\underline{\sf{k=800\ J\,m^{-2}}}}$