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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 l0, and it is stretched (or compressed) to a length l0 + x. then it exerts a force
Fspring=kx, ....(1)
where k is the spring constant in [N/m], and the force of the spring is opposite the direction of the pull (or push). (See Figure 11-1.)
Let's be clear about the forces involved here. Figure 11-2 shows a spring attached to a wall stretched by a hand. Figure 11-3 shows all the forces involved. The force F1 is the force the spring exerts on the wall, and F2 is the force the wall exerts on the spring. These forces are equal in magnitude because of the third law of motion. The force F3 is the force the hand exerts on the spring, and F4 is the force the spring exerts on the hand. The forces F3 and F4 are equal in magnitude because of the third law of motion as well. The forces F2 and F3 add to zero because of the second law of motion, because the spring is not accelerating, giving force balance. All these forces are equal in magnitude (for various reasons), but there are four forces.


In spring problems, as in many types of physics problems, it is important to follow the energy. The potential energy of a stretched or compressed spring is as follows:
This energy is the second type of potential energy we have encountered. In many problems, we can treat this potential energy like gravitational potential energy.
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
EK + EP = constant ...(3)
The expression EP includes both gravitational and spring energies, although in most problems it represents just one or the other.

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 = FspringΔxcosφ = FspringΔx = (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 11-4, showing the system at three different times.) In the work equation we assume the force is constant. What shall we do?


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Figure 11-4


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,


Wtot = ΔEK


But the kinetic energy of the mass comes from the potential energy of the spring, so we have


Wtot = ΔEK = –ΔEP


EP1 – EP2


= 1/2 kx2 – 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.


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