Question
There 46.5J of elastic potential energy stored in a spring. How far is the spring stretched if the spring constant is 455N/m
Answers
Chris
PE = (1/2)kx^2
(46.5) = (1/2)(455)(x)^2
x = 0.452 meters
(46.5) = (1/2)(455)(x)^2
x = 0.452 meters