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Question

A force $$F_{x}$$ acts on a particle such that its position $$x$$ changes as shown in figure.
The work done by the particle as it moves from $$x = 0$$ to $$20\ m$$ is
671463_c83bf09a81ba4b8092025f9c24d1f836.png


A
37.5 J
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B
10 J
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C
15 J
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D
22.5 J
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E
45 J
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Solution

The correct option is E $$45\ J$$
Key Concept As, we know, work done by a variable force, $$W_{n_{i}\rightarrow n_{t}}$$
$$=$$ area under the force $$-$$ displacement curve.
Work $$=$$ Area of $$\triangle ABF +$$ Area of $$\triangle BCEF +$$ Area of $$\triangle ECD$$
Work done by the particle as it moves from $$n = 0$$ to $$20\ m$$,
$$W = \dfrac {1}{2} \times 3\times 5 + 10\times 3 + \dfrac {1}{2} \times 5\times 3$$
$$\Rightarrow W = \dfrac {15}{2} + 30 + \dfrac {15}{2}$$
$$\Rightarrow W = 15 + 30 = 45\ J$$.
707714_671463_ans_2fdb86b2eaca4c858f1ab7d85bd42de4.jpg

Physics

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