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5.Work, Energy, Power and Collision
normal
The work done by a force $\vec F = (-6x^3\hat i)\, N$, in displacing a particle from $x = 4\, m$ to $x = -2\, m$ is .............. $\mathrm{J}$
A
$360$
B
$240$
C
$-240$
D
$-360$
Solution
$\mathrm{W}=\int_{\mathrm{A}}^{\mathrm{B}} \mathrm{F}_{\mathrm{x}} \mathrm{dx} \Rightarrow \mathrm{W}=\int_{\mathrm{x}-4}^{\mathrm{x}-2}\left(-6 \mathrm{x}^{3}\right) \mathrm{d} \mathrm{x}$
$=-6\left[\frac{\mathrm{x}^{4}}{4}\right]_{x-4}^{x-2}=\left(\frac{-3}{2}\right)(-240)=360 \mathrm{J}$
Standard 11
Physics
