Question

A block of mass m moving at a speed ν compresses a spring through a distance x before its speed is halved. Find the spring constant of the spring.

Answer 1

Let the velocity of the body at P be ν.

So, the velocity of the body at Q is ν2.

Energy at point P = Energy at point Q

So, 12mν2P−(12) mv2Q=12kx2⇒12kx2=12m (V2P−V2Q)⇒kx2= m (ν2−ν24)⇒kx2=m (4ν2−ν2)4⇒k=3mv24x2

Hope it helps...!!!

Answer 2

The spring constant of the spring is $\left(\frac{3 m v^{2}}{4 x^{2}}\right)$

Explanation:  

Step 1:

The initial velocity is considered as $V$.

k is considered as the spring constant of the spring

After compression the velocity changes = $v / 2$

Here, a block's initial kinetic energy = $\frac{1}{2} m v^{2}$

After spring compression,

Step 2:

Total energy at the point x = Kinetic energy of a block + Potential Spring Energy that contained.  

$\frac{1}{2} m v^{2}=\frac{1}{2} m\left(\frac{v}{2}\right)^{2}+\frac{1}{2} k x^{2}$

$\frac{1}{2} k x^{2}=\frac{1}{2} m v^{2}-\frac{1}{2} m\left(\frac{v}{2}\right)^{2}$

$\frac{1}{2} k x^{2}=\frac{1}{2}\left(m v^{2}-m\left(\frac{v}{2}\right)^{2}\right)$

Step 3:

$\frac{1}{2}$ common both side and cancel each other  

$k x^{2}=\left(m v^{2}-m\left(\frac{v}{2}\right)^{2}\right)$

$k x^{2}=m\left(v^{2}-\left(\frac{v}{2}\right)^{2}\right)$

$k x^{2}=m\left(v^{2}-\frac{v^{2}}{4}\right)$

$k x^{2}=m\left(\frac{4 v^{2}-v^{2}}{4}\right)$

$k x^{2}=m\left(\frac{3 v^{2}}{4}\right)$

$k=\left(\frac{3 m v^{2}}{4 x^{2}}\right)$

Then k (a constant spring) is equal to = $\left(\frac{3 m v^{2}}{4 x^{2}}\right)$