Question

A block of mass 0.252 kg is placed on top of a light, vertical spring of force constant 4 825 N/m and pushed downward so that the spring is compressed by 0.105 m. After the block is released from rest, it travels upward and then leaves the spring. To what maximum height above the point of release does it rise?

Answer 1

Answer:

10.77m

Explanation:

The elastic potential energy of the spring when compressed with the mass is converted to the kinetic energy of the mass when released. This ie expressed in the following equation;

$\frac{1}{2}ke^2=\frac{1}{2}mu^2..............(1)$

where k is the force constant of the spring, e is the compressed length, m is the mass of the block and u is the velocity with which the block leaves the spring after being released.

If we make u the subject of formula from equation (1) we obtain the following;

$u=\sqrt{\frac{ke^2}{m}}................(2)$

Given;

e = 0.105m,

k = 4825N/m,

m = 0.252kg,

u = ?

Substituting all values into equation (2) we obtain the following;

$u=\sqrt{\frac{4825*0.105^2}{0.252}}................(2)\\u=14.53m/s$

The maximum height attained is then obtained from the third equation of motion as follows, taking g as $9.8m/s^2$

$v^2=u^2-2gh.........(3)$

v = 0m/s

Hence

$h=\frac{u^2}{2g}\\h=\frac{14.53^2}{2*9.8}\\h= 10.77m$