Question

A body of mass $0.1\text{kg}$, moving with a velocity of $10\text{m/s}$ hits a horizontal spring (fixed at the one end) of force constant $1000\text{N/m}$ and comes to rest after compressing the spring. The compression of the spring is,

• A. $0.01\text{m}$
• B. $0.1\text{m}$
• C. $0.2\text{m}$
• D. $0.5\text{m}$

Answer 1

The correct option is B $0.1\text{m}$

Decrease in kinetic energy of the body $=$ Increase in elastic potential energy of the spring.
$\therefore \frac{1}{2}m{v}^2=\frac{1}{2}k{x}^2$

or $x=v\sqrt{\frac{m}{k}}=10\times \sqrt{\frac{0.1}{1000}}=0.1\text{m}$

Hence, $\left(B\right)$ is the correct answer.

$$\begin{array}{c}\text{Why this questrion ?}\\ \text{If work done by net external force acting on the system is zero, then mechanical energy is conserved}{k}_i+U_i=K_f+U_f\text{.}\end{array}$$