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Sagot :
A 0. 20-kg ball is attached to a vertical spring. The spring constant is 28 n/m. when released from rest, the ball will fall at a distance of x = 0.14 m before being brought to a momentary stop by the spring.
The spring constant is the force needed to stretch or compress a spring, divided by the distance that the spring gets longer or shorter. It's used to determine stability or instability in a spring
Elastic potential energy is Potential energy stored as a result of deformation of an elastic object, such as the stretching of a spring. It is equal to the work done to stretch the spring, which depends upon the spring constant k as well as the distance stretched.
When ball is released from rest with spring , it will loose some potential energy . Hence , loss in potential energy of the ball = gain in potential energy of spring
m*g*x = 1/2 * k * [tex]x^{2}[/tex]
0.20 * 9.8 * x = 1/2 * 28 * [tex]x^{2}[/tex]
x = (2 * 0.20 * 9.8) / 28
x = 0.14 m
To learn more about spring constant here
https://brainly.com/question/14670501
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