Answer:
The height at point of release is 10.20 m
Explanation:
Given:
Spring constant [tex]k = 5 \times 10^{3} \frac{N}{m}[/tex]
Compression [tex]x = 0.10[/tex] m
Mass of block [tex]m = 0.250[/tex] kg
Here spring potential energy converted into potential energy,
[tex]mgh = \frac{1}{2} kx^{2}[/tex]
For finding at what height it rise,
[tex]0.250 \times 9.8 \times h = \frac{1}{2} \times 5 \times 10^{3} \times (0.10) ^{2}[/tex] ( ∵ [tex]g = 9.8 \frac{m}{s^{2} }[/tex] )
[tex]h = 10.20[/tex] m
Therefore, the height at point of release is 10.20 m
This question involves the concepts of the law of conservation of energy, elastic potential energy, and gravitational potential energy.
The block will rise "10.2 m" high above the point of its release.
According to the law of conservation of energy, the elastic potential energy stored by spring must be equal to the gravitational potential energy acquired by the block.
[tex]mgh = \frac{1}{2}kx^2[/tex]
where,
m = mass = 0.25 kg
g = acceleration due to gravity = 9.81 m/s²
h = heoght = ?
k = spring constant = 5000 N/m
x = compression = 0.1 m
Therefore,
[tex]h=\frac{(5000\ N/m)(0.1\ m)^2}{2(0.25\ kg)(9.81\ m/s^2)}[/tex]
h = 10.2 m
Learn more about the law of conservation of energy here:
brainly.com/question/20971995?referrer=searchResults
The attached picture explains the law of conservation of energy.
