Questions

A particle of mass 2 kg is moved along a line $3y=4x+10$ under the influence of a force $\overrightarrow{\mathrm{F}}=(4\widehat{\mathrm{i}}-3\widehat{\mathrm{j}})\mathrm{N}$ . Find the work done by the force between the points $\left(-\frac{10}{4},0\right)$ and $\left(0,\frac{10}{3}\right)$

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a

20 J

b

30 J

c

36.5 J

d

Zero

detailed solution

Correct option is D

The particle is moved along the line 3y=4x+10,y=43x+103=mx+c , . Hence the slope of this line is 4/3. Again for the force acting slope=tanθ=FyFx=−34 . So product of the slopes= -1. We can say that the two lines are perpendicular to each other. W=F.S.Cosθ=F.S.Cos900=0

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