The work done by a force F=−6x3i^N in displacing a particle from x = 4 m to x = -2 m is

dW=F⋅ds=−6x3i^⋅(dxi^+dyj^+dzk^)or dW=−6x3⋅dx∴ W=∫4−2 −6x3⋅dx=−32x44−2 =−32[16−256]=360J

The work done by a force F=−6x3i^N in displacing a particle from x = 4 m to x = -2 m is

dW=F⋅ds=−6x3i^⋅(dxi^+dyj^+dzk^)or dW=−6x3⋅dx∴ W=∫4−2 −6x3⋅dx=−32x44−2 =−32[16−256]=360J

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The work done by a force $F = (- 6 x^{3} \hat{i}) N$ in displacing a particle from x = 4 m to x = -2 m is

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will depend upon the path

420 J

360 J

-240 J

answer is B.

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Detailed Solution

$\begin{array}{l} dW = F \cdot ds = (- 6 x^{3} \hat{i}) \cdot (dx \hat{i} + dy \hat{j} + dz \hat{k}) \\ o r dW = - 6 x^{3} \cdot dx \\ ∴ W = \int_{4}^{- 2} - 6 x^{3} \cdot dx = \frac{- 3}{2} {[x^{4}]}_{4}^{- 2} \\ = - \frac{3}{2} [16 - 256] = 360 J \end{array}$