posted by victoria on .
A spring (spring 1) with a spring constant of 520N/m is attached to a wall and connected to another weaker spring (spring 2) with a spring constant of 270N/m on a horizontal surface. Then an external force of 70N is applied to the end of the weaker spring (#2).
how much potential energy is stored in each spring
F = k1*x1
70 = (520)*x1
F = k2*x2
70 = 270*x2
Plug x's into potential energy for a spring equation U = 0.5kx^2 for respective springs
70N * 1m/520N. = 0.1346 m.
PE=0.5F * d = 0.5*70 * 0.1346 = 4.71 J.
70N/270N * 1m. = 0.2593 m.
PE=0.5F * d = 0.5*70 * 0.2593 = 9.07 J.