22. A spring is stretched 0.418 m from equilibrium by a
force of magnitude 1.00 102 N.
(a) What is the force constant of the spring?
(b) What is the magnitude of force required to
stretch the spring 0.150 m from equilibrium?
(c) How much work must be done on the spring to
stretch it 0.150 m from equilibrium and to compress
it 0.300 m from equilibrium?

Question

22. A spring is stretched 0.418 m from equilibrium by a
force of magnitude 1.00  102 N.
(a) What is the force constant of the spring?
(b) What is the magnitude of force required to
stretch the spring 0.150 m from equilibrium?
(c) How much work must be done on the spring to
stretch it 0.150 m from equilibrium and to compress
it 0.300 m from equilibrium?

anonymous · Answer

$$F = ks$$ where k is the spring constant and s is the distance stretched. (+ positive Tension, - negative compression) so: $$102N = 0.418k$$ means that $$k = 102/0.418 = 244 $$

b) $$F = 244 * 0.150 = 36.6$$

c) $$W = Fd = ks^2$$ so $$W_1 = 244*.150^2 = 5.5 J$$ and $$W_2 = 244 * (-0.300)^2 = 21.96 = 22 J$$