The work done by a force vec F, = ,( - ,6{x^3 | vedclass

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Question

$\mathrm{W}=\int_{\mathrm{A}}^{\mathrm{B}} \mathrm{F}_{\mathrm{x}} \mathrm{d} \mathrm{x} \Rightarrow \mathrm{W}=\int_{\mathrm{x}-4}^{\mathrm{x}-2}\lef

Vedclass · Accepted Answer

$360$

Vedclass · Answer

$\mathrm{W}=\int_{\mathrm{A}}^{\mathrm{B}} \mathrm{F}_{\mathrm{x}} \mathrm{d} \mathrm{x} \Rightarrow \mathrm{W}=\int_{\mathrm{x}-4}^{\mathrm{x}-2}\left(-6 \mathrm{x}^{3}\right) \mathrm{d} \mathrm{x}$

$=-6\left[\frac{x^{4}}{4}\right]_{x-4}^{x-2}=\left(\frac{-3}{2}\right)(-240)=360 \mathrm{J}$

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

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