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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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Most popular questions from this chapter

When liquid medicine of density \(\mathrm{S}\) is to be put in the eye. It is done with the help of a dropper as the bulb on the top of the dropper is pressed a drop forms at the opening of the dropper we wish to estimate the size of the drop. We dirst assume that the drop formed at the opening is spherical because the requires a minimum increase in its surface energy. To determine the size we calculate the net vertical force due to surface tension \(\mathrm{T}\) when the radius of the drop is \(\mathrm{R}\). When this force becomes smaller than the weight of the drop the drop gets detached from the dropper. If $\mathrm{r}=5 \times 10^{-4} \mathrm{~m}, \mathrm{p}=10^{3} \mathrm{~kg} \mathrm{~m}^{-3}=10 \mathrm{~ms}^{-2} \mathrm{~T}=0.11 \mathrm{~N} \mathrm{~m}^{-1}$ the radius of the drop when it detaches from the dropper is approximately (A) \(1.4 \times 10^{-3} \mathrm{~m}\) (B) \(3.3 \times 10^{-3} \mathrm{~m}\) (C) \(2.0 \times 10^{-3} \mathrm{~m}\) (D) \(4.1 \times 10^{-3} \mathrm{~m}\)

A tank is filled with water up to a height \(\mathrm{H}\). Water is allowed to come out of a hole P in one of the walls at a depth \(\mathrm{D}\) below the surface of water express the horizontal distance \(\mathrm{x}\) in terms of \(\mathrm{H}\) and \(\mathrm{D}\). (B) $\left.\mathrm{x}={ }^{\alpha} \sqrt{[}\\{\mathrm{D}(\mathrm{H}-\mathrm{D})\\} / 2\right]$ (D) \(\mathrm{x}=4[\mathrm{D}(\mathrm{H}-\mathrm{D})]\)

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If a rubber ball is taken at the depth of \(200 \mathrm{~m}\) in a pool, Its volume decreases by \(0.1 \%\). If the density of the water is $1 \times 10^{3}\left(\mathrm{~kg} / \mathrm{m}^{3}\right) \& \mathrm{~g}=10\left(\mathrm{~m} / \mathrm{s}^{2}\right)$. Then what will be the volume elasticity in ? (A) \(10^{8}\) (B) \(2 \times 10^{8}\) (C) \(10^{9}\) (D) \(2 \times 10^{9}\)

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