Question

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 the figure. As a result, the spring is compressed by an amount x = 10 cm. What is the force constant of the spring? Take $g=10m{s}^{-2}$.

• A. $600N{m}^{-1}$
• B. $800N{m}^{-1}$
• C. $1000N{m}^{-1}$
• D. $1200N{m}^{-1}$

Answer 1

The correct option is C

$1000N{m}^{-1}$

Since the platform is depressed by an amount x, the total change in potential energy of the body is mg (h + x). This work is stored in the spring in the form of elastic potential energy.
$\therefore \frac{1}{2}k{x}^2=mg(h+x)\Rightarrow k=\frac{2mg(h+x)}{x^2}$
Given, h = 0.4 m, x = 0.1 m, m = 1 kg and $g=10m{s}^{-2}$. Substituting these values, we get $k=1000N{m}^{-1}$. Hence, the correct choice is (c).