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Chapter 7: Problem 916

\(\mathrm{k}\) is the force constant of a spring what will be the work done in increasing its extension form \(\ell_{1}\) to \(\ell_{2}\) be ? (A) \(\mathrm{k}\left(\ell-\ell_{2}\right)\) (B) \((\mathrm{k} / 2)\left(\ell_{2}+\ell_{1}\right)\) (C) \(\mathrm{k}\left(\ell_{2}^{2}+\ell_{1}^{2}\right)\) (D) \((\mathrm{k} / 2)\left({ }_{2}^{2}-\ell_{1}^{2}\right)\)

Short Answer

Expert verified
The short answer is: \(W = \frac{k}{2}(\ell_{2}^{2} - \ell_{1}^{2})\).

Step by step solution

01

Recall Hooke's Law

Hooke's Law states that the force exerted by a spring is directly proportional to its displacement from the equilibrium position, represented by -kx, where k is the force constant of the spring and x is the displacement. The equation for Hooke's Law is: \[F = -kx\]
02

Calculate work done using the work-energy theorem.

The work-energy theorem states that the amount of work done on an object is equal to the change in its potential energy. For a spring, potential energy is given by the equation: \[U = \frac{1}{2}kx^2\] So, the change in potential energy by changing the extension from \(\ell_1\) to \(\ell_2\) is given by: \[\Delta U = U_{\ell_{2}} - U_{\ell_{1}} = \frac{1}{2}k\ell_{2}^{2} - \frac{1}{2}k\ell_{1}^{2}\] Since work done is equal to the change in potential energy, we have: \[W = \Delta U\]
03

Substitute the values for potential energy change into the work done equation and simplify.

Combining the expressions for potential energy change and work done, we have: \[W = \frac{1}{2}k\ell_{2}^{2} - \frac{1}{2}k\ell_{1}^{2}\] Factor out \(\frac{1}{2}k\) from both terms to simplify the expression: \[W = \frac{k}{2}(\ell_{2}^{2} - \ell_{1}^{2})\]
04

Compare to given choices.

Comparing the simplified expression for work done to the given choices, we can see that it matches option (D): (D) \((\mathrm{k} / 2)\left({ }_{2}^{2}-\ell_{1}^{2}\right)\) Thus, the correct choice is option (D).

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