Question

A block of mass $m$ moving with speed v compresses a spring through distance $x$ before is halved. What is the value of spring constant?

• A. $\frac{3m{v}^2}{4x^2}$
• B. $\frac{m{v}^2}{4x^2}$
• C. $\frac{m{v}^2}{2x^2}$
• D. $\frac{2m{v}^2}{x^2}$

Answer 1

The correct option is A $\frac{3m{v}^2}{4x^2}$

Let the velocity at starting is $v$.

After compression change in velocity $=\frac{v}{2}$

Here, Initial kinetic energy of a block $=\left(\frac{1}{2}\right)m{v}^2$

After compression of spring,

Total energy at the point $x$= Kinetic energy of a block +Potential Energy which stored in the spring.

$\begin{array}{c}\left(\frac{1}{2}\right)m{v}^2=\frac{1}{2}m{\left(\frac{v}{2}\right)}^2+\frac{1}{2}k{v}^2\\ \frac{1}{2}k{v}^2=\frac{1}{2}m{\left(\frac{v}{2}\right)}^2-\frac{1}{2}\left(m{v}^2\right)\\ k{x}^2=m\left(v^2-\frac{v^2}{4}\right)\\ k{x}^2=\frac{3m{v}^{{}^2}}{4}\\ \therefore k=\frac{3m{v}^{{}^2}}{4x^2}\end{array}$