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

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

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 $$20\ m$$,$$W = \dfrac {1}{2} \times 3\times 5 + 10\times 3 + \dfrac {1}{2} \times 5\times 3$$$$\Rightarrow W = \dfrac {15}{2} + 30 + \dfrac {15}{2}$$$$\Rightarrow W = 15 + 30 = 45\ J$$.Physics

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