Question

A spring of spring constant $50\text{N/m}$ initially at its natural length is compressed through $1\text{cm}$. Find the work done by the spring-force on the external agent, compressing the spring.

• A. $-2.5\text{mJ}$
• B. $2.5\text{mJ}$
• C. $3.5\text{mJ}$
• D. $-3.5\text{mJ}$

Answer 1

The correct option is A $-2.5\text{mJ}$

The spring force is the restoring force given by $F=-kx$ and it will act in a direction opposite to the displacement of the point on the spring.
Therefore, work done by spring force,
$W_{\text{spring}}=\int F_{\text{spring}}dx$
$W_{\text{spring}}={\int }_0^x-kxdx=-\frac{1}{2}k{x}^2$
$\Rightarrow W=-\frac{1}{2}\times \left(50\text{N/m}\right)\times (0.01\text{m}{)}^2$
$\text{Or},W=-2.5\times {10}^{-3}\text{J}=-2.5\text{mJ}$

The force will be opposite to the displacement of the point of application, hence it is evident that work done by spring-force will be negative.

$\underset{–}{\text{Alternative solution}:}$

Spring force is a conservative force, hence we can write
$W_{\text{spring}}=-\Delta U$
${}^{\prime }{U}^{\prime }$ represents the spring potential energy
When spring was at its natural length, $U_i=0$
After compressing the spring through distance $x$,
$U_f=\frac{1}{2}k{x}^2$
$\Rightarrow W_{\text{spring}}=-(U_f-U_i)=-U_f$
$\text{Or},W_{\text{spring}}=-\frac{1}{2}k{x}^2$
$\text{Or},W_{\text{spring}}=-\frac{1}{2}\times \left(50\text{N/m}\right)\times (0.01\text{m}{)}^2$
$\therefore W_{\text{spring}}=-2.5\times {10}^{-3}\text{J}=-2.5\text{mJ}$