Question

Consider the force exerted by a spring that obeys Hooke's law. Find U(xf​)−U(x0​)=−∫x0​xf​​Fs​⋅ds where Fs​=−kxi^,ds=dxi^ and the spring constant k is positive. Express your answer in terms of k,x0​, and xf​.

Answer 1

The expression U(xf​)−U(x0​)=−∫x0​xf​​Fs​⋅ds describes the potential energy stored in a spring according to Hooke's Law. The change in potential energy, U(xf​)−U(x0​), when the spring is stretched or compressed from an initial position x0 to a final position xf, is equal to k/2 * (xf² - x0²), where k is the spring constant.

Step-by-step explanation:

The expression U(xf​)−U(x0​)=−∫x0​xf​​Fs​⋅ds represents the potential energy stored in the spring at a certain stretch or compression. Here, U(xf) is the potential energy at the final position, U(x0) is the potential energy at the initial position, Fs is the force exerted by the spring, k is the spring constant, and xi and xf are the initial and final positions.

The force exerted by the spring can be represented by Hooke's law, which states that the force exerted by the spring is proportional to the distance it is stretched or compressed, and is given by Fs = -kx, where x is the displacement of the spring from its equilibrium position.

To find U(xf​)−U(x0​), we need to integrate Fs from x0 to xf. Since Fs = -kx, the integral becomes -∫x0​xf​​(-kx)dx, or k/2[xf² - x0²]. Therefore, the change in the potential energy of the spring is represented by U(xf​)−U(x0​) = k/2 (xf² - x0²).

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