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# The amount of work done in stretching a spring from a stretched length of $10 \mathrm{cm}$ to a stretched length of $20 \mathrm{cm}$ is-(A) Equal to the work done in stretching it from $20 \mathrm{cm}$ to $30 \mathrm{cm}$(B)Less than the work done in stretching it from $20 \mathrm{cm}$ to $30 \mathrm{cm}$.(C) More than the work done in stretching it from $20 \mathrm{cm}$ to $30 \mathrm{cm}$.(D) Equal to the work done in stretching it from 0 to $10 \mathrm{cm}$.

Last updated date: 20th Sep 2024
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Hint: We know that work can be either positive or negative: if the force has a component in the same direction as the displacement of the object, the force is doing positive work. If the force has a component in the direction opposite to the displacement, the force does negative work. One joule is defined as the amount of work done when a force of one newton is exerted through a distance of one meter. In the English system of units, where force is measured in pounds, work is measured in a unit called the foot-pound. When a force acts (pushes or pulls) on an object, it changes the object's speed or direction (in other words, makes it accelerate). The bigger the force, the more the object accelerates. When a force acts on an object, there's an equal force (called a reaction) acting in the opposite direction.

Work done in stretching a spring to $x$ length is $\mathrm{W}=\dfrac{1}{2} \mathrm{kx}^{2}$
$\therefore$ Work one in stretching from $\mathrm{x}_{1}$ to $\mathrm{x}_{2}$ is proportional to $W\propto x_{2}^{2}-x_{1}^{2}$
hence $\dfrac{\mathrm{W}_{20,30}}{\mathrm{W}_{10,20}}=\dfrac{500}{300}$