91Ó°ÊÓ

The given data can be listed below,

• The mass of the block is,$m=4\text{ k²µ}$.
• The force constant of the spring is,$k=200\text{ N/³¾}$.
• The compressed length of the spring is,$x_1=0.025\text{″¾}$.

The well-known equation used to describe a mass on a spring is Hooke's law. It measures tension in a solid within the elastic limit.

The work done on the blockby spring is given by,

$W=\frac{1}{2}k{x}_1^2−\frac{1}{2}k{x}_2^2$

Here, k is the spring constant,$x_1$is the compressed distance, and$x_2$is the final compressed distance whose value is zero as it eventually comes to rest.

Substitute all the values in the above,

$\begin{array}{c}W=\frac{1}{2}\left(200\text{ N/³¾}\right){\left(0.025\text{″¾}\right)}^2−\frac{1}{2}\left(200\text{ N/³¾}\right){\left(0\text{″¾}\right)}^2\\ =\left(100\text{ N/³¾}\right){(−0.025\text{″¾})}^2\\ =0.0625\text{ J}\end{array}$

Thus, the work done on the block by spring from its initial position to where it comes to its original position is $0.0625\text{ J}$.

By the work-energy theorem, the velocity of the block is given by,

$\begin{array}{c}W=K_f−K_i\\ =\frac{1}{2}m{v}_2^2−\frac{1}{2}m{v}_1^2\\ v_2=\sqrt{\frac{2W}{m}}\end{array}$

Here, m is the mass of the block, $v_1$is the initial velocity of the block, and $v_2$is the final velocity of the block.

Substitute all the values in the above,

$\begin{array}{c}W=\sqrt{\frac{2×0.0625\text{ J}}{4\text{ k²µ}}}\\ =0.18\text{″¾/s}\end{array}$

Thus, the velocity of the block after it leaves the spring is$0.18\text{″¾/s}$ .