One end of a spring is fixed, while the other can be moved in such a way to stre

Question

One end of a spring is fixed, while the other can be moved in such a way to stretch or compress the spring. Suppose we choose the origin to be the position of the left end of the cart when the system is at equilibrium, and the direction such that a displacement that stretches the spring is positive, as seen in Fig. 11.6, (where a 0.75 kg cart has been attached to the free end of the spring).

2. The cart is moved to the right by 2.0 cm, as shown in Fig. 11.6. WHat is the elastic potential energy stored in the spring? Does the energy depend on the mass of the cart?

Explanation / Answer

The spring constant can be calculated by finding the slope of this graph.

Spring force equation
Fs = -kx

We should expect the slope of this graph to be equal to the negative of the spring constant.

slope of the graph is:
(-10-10)/(15-(-15)) = -0.667

So the spring constant is 0.667N/cm or 66.7N/m

Elastic potential can be calculated by:

U = 1/2kx^2

x is 2 cm or 0.02m so:

U = 1/2(66.7N/m)(0.02m)^2 = 0.0133J

Which does not depend on the mass of the cart.

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