Springs
If you stretch a spring, the spring exerts a pull on you. The more you stretch it, the more it pulls. If you compress a spring capable of being compressed, then it exerts a push. Hooke's law states that the force is proportional to the extension or compression.Hooke's Law If a spring has resting length l_{0}, and it is stretched (or compressed) to a length l_{0} + x. then it exerts a force

â€¦(2) 
Conservation of Energy, Again
If there is no friction, no crashing, no heat generation, and if each force is a potential force or does no work, then the sum of kinetic and potential energies is constant, so we have

One end of a horizontal spring (k = 20 N/m) is connected to a wall, and the other end is connected to a mass m (0.2 kg). The spring is compressed 0.1 m from equilibrium. After it is released, how much work does the spring do in order to push the mass to the equilibrium position?
Our first idea is to write the work equation: W = F_{spring}Î”xcosÏ† = F_{spring}Î”x = (20 N/m 3 0.1 m) (0.1 m) = 0.2 J. But this would be WRONG. The force of the spring begins at (20 N/m) (0.1 m) = 2 N, but then it decreases as the spring moves the mass. (See Figure 114, showing the system at three different times.) In the work equation we assume the force is constant. What shall we do?
Figure 114
Let's think about the flow of energy after the release. The work done by the spring on the mass is the same as the change in kinetic energy of the mass (by energy conservation), that is,
W_{tot} = Î”E_{K}
But the kinetic energy of the mass comes from the potential energy of the spring, so we have
W_{tot} = Î”E_{K} = â€“Î”E_{P}
= E_{P1} â€“ E_{P2}
= 1/2 kx^{2} â€“ 0
= 1/2 (20 N/m)(0.1 m)^{2} = 0.1 J
This is one of those problems in which blind plugging into the formulas is to no avail, but thinking about the energy flow is the key to success.