## Illustration 6.4: Springs

y = m | spring constant, k = N/m

The fact that the spring force varies with position means that while we can determine the force, we cannot determine the velocity of an object attached to a stretched spring using kinematic equations for constant acceleration. Why? The force is not constant (it varies with position) and therefore the acceleration is not constant. What can we do? We can use the work-energy theorem.

The spring-ball system shown in the animation can be stretched by click-dragging the 1-kg dark blue ball (position is given in meters and time is given in seconds). The black arrow attached to the ball shows the net, i.e., total, force on the ball. The pale blue ball on the left is the free-body diagram for the dark blue ball. The red and green arrows attached to the pale blue ball show the spring and gravitational forces, respectively. The acceleration due to gravity is 9.8 m/s2 in this animation. Restart.

Hooke's law states that the force that the spring exerts is F = -k x, where k is the spring constant and x is measured from the equilibrium position of the spring. In this Illustration the initial position of the spring and the spring constant can be changed by using the text boxes.

So how do we determine the work done by the spring? We need to calculate the integral of Fcos(θ) Δx, where F and Δx are the magnitude of the force vector and the displacement vector, respectively. We must calculate the integral because the force is not a constant.

Consider k = 2 N/m and y = 5 m and run the animation. Initially the spring is compressed and the net force points down and the infinitesimal displacement points down; therefore, cos(θ) = 1. We determine that the work done by the spring is initially positive, yielding an increasing kinetic energy. After passing through equilibrium (y = 0.1 m for this k), however, the net force is now upward, while the infinitesimal displacement still points down. Thus cos(θ) = -1, and the work is negative. Therefore, the kinetic energy decreases until the spring is maximally extended and v = 0 m/s. The process then reverses with a positive amount of work until the mass again passes through equilibrium and the spring does a negative amount of work until the mass is at rest at its starting position y = 5 m. The process repeats indefinitely if there are no resistive forces.

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