Question

a force of 10 N holds an ideal spring with a 20-N/m spring constant in compression. The potential energy stored in the spring is..? 0.5J, 2.5J, 5.0J, 10J, or 200J?

POtential energy? Wouldn't it be 1/2 kx^2,


but x= F/k, so 1/2 kx^2= 1/2 k*F^2/k^2 and that can be reduced.

Answer 1

A 0.5-kg block attached to an ideal spring with a spring constant of 65N/m oscillates on a
horizontal frictionless surface. The total mechanical energy is 0.79 J. What is the greatest extension
of the spring from its equilibrium length?

Answer 2

square root of(.79/(.5*65))

Answer 3

a. PE = 1/2 * kx^2.

or d = 10 * 1m/20N = 0.5 m.
PE = 0.5*F*d = 0.5 * 10 * 0.5 = 2.5 J.

Answer 4

F=10N
k=20N/m
x=?

PE=(1/2)(k)(x^2)
x=F/k=10/20=0.50

PE=1/2(20)(0.50)^2
=2.5J