Question

An object of mass 2.5 kg sliding on a horizontal frictionless surface with a speed of 4m / s collides with a horizontal relaxed spring as shown in the figure . If the spring is compressed by a distance of 40 cm and the block comes to rest , then the spring constant k is equal to :

Answer 1

Given:

The mass of the object is,

$m=2.5\text{ kg}$

The speed of the object is,

$v=4\text{ m/s}$

The compression of the spring is,

$\begin{gathered} x=40\text{ cm} \\ =0.40\text{ m} \end{gathered}$

To find:

The spring constant

Explanation:

The kinetic energy of the box provides the potential energy of the spring.

If the spring constant of the spring be 'k', the potential energy is,

$\frac{1}{2}kx^2$

The kinetic energy of the mass is,

$\frac{1}{2}mv^2$

Now we can write,

$\begin{gathered} \frac{1}{2}kx^2=\frac{1}{2}mv^2 \\ k=m\frac{v^2}{x^2} \end{gathered}$

Substituting the values we get,

$\begin{gathered} k=2.5\times\frac{4\times4}{0.40\times0.40} \\ =250\text{ N/m} \end{gathered}$

Hence, the spring constant is 250 N/m.