A spring has a force constant of 590.0 N/m.
(a) Determine the potential energy stored in the spring when the spring is stretched 3.77 cm from equilibrium.
(b) Determine the potential energy stored in the spring when the spring is stretched 3.26 cm from equilibrium.
(c) Determine the potential energy stored in the spring when the spring is unstretched.
2 answers
see other post.
a. F = 590N/m * 0.0377m = 22.24 N.
PE=0.5F * d = 0.5*22.24 * 3.77 = 247 J.
b. Same procedure as part "a".
c. PE = 0.5F * d = 0.5F * 0 = 0.
PE=0.5F * d = 0.5*22.24 * 3.77 = 247 J.
b. Same procedure as part "a".
c. PE = 0.5F * d = 0.5F * 0 = 0.
1 answer