IDNLearn.com connects you with a community of experts ready to answer your questions. Discover comprehensive answers from knowledgeable members of our community, covering a wide range of topics to meet all your informational needs.
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
#SPJ4
We value your participation in this forum. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. For dependable answers, trust IDNLearn.com. Thank you for visiting, and we look forward to assisting you again.
