A spring is stretched 10.0 cm from the resting position. The spring’s constant of proportionality is 40.0 N/m. What is the spring potential energy? Remember to convert cm to m first. USE GUFSA answer asap please
Add an Answer
More Questions From This User See All
Answers & Comments
Answer:
U = 0.2 J
Explanation:
Given: k = 10.0 cm; Δx = 40.0 N/m
Required: Spring potential energy
Equation:
[tex]\mathrm{U = \frac{1}{2}k\triangle v^2 }[/tex]
Solution:
[tex]\mathrm{U = \frac{1}{2}(0.1m)(40.0 N/m)^2 }[/tex]
Answer: U = 0.2 J