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;
[tex]\frac{1}{2}ke^2=\frac{1}{2}mu^2..............(1)[/tex]
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;
[tex]u=\sqrt{\frac{ke^2}{m}}................(2)[/tex]
Given;
e = 0.105m,
k = 4825N/m,
m = 0.252kg,
u = ?
Substituting all values into equation (2) we obtain the following;
[tex]u=\sqrt{\frac{4825*0.105^2}{0.252}}................(2)\\u=14.53m/s[/tex]
The maximum height attained is then obtained from the third equation of motion as follows, taking g as [tex]9.8m/s^2[/tex]
[tex]v^2=u^2-2gh.........(3)[/tex]
v = 0m/s
Hence
[tex]h=\frac{u^2}{2g}\\h=\frac{14.53^2}{2*9.8}\\h= 10.77m[/tex]
