A force F=a+bx acts on a particle in the x-direction where a and b are constants. Find the work done by this force during a displacement from x=0 to x=d.

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A force F=−kx2(x≠0) acts on a particle in x−direction. Find the work done by this force in displacing the particle from x=+a to x=+2a. Here k is a positive constant.

The correct option is C −k2aBy definition work done by force is W=∫Fdx =−k∫2aadxx2 =−k[x−2+1−2+1]2aa =k[1x]2aa =k[12a−1a] W=−k2a

A force $F = - \frac{k}{x^{2}} (x \neq 0)$ acts on a particle in $x -$ direction. Find the work done by this force in displacing the particle from $x = + a$ to $x = + 2 a .$ Here $k$ is a positive constant.

The correct option is C $- \frac{k}{2 a}$
By definition work done by force is
$W = \int F d x$
$= - k \int_{a}^{2 a} \frac{d x}{x^{2}}$
$= - k {[\frac{x^{- 2 + 1}}{- 2 + 1}]}_{a}^{2 a}$
$= k {[\frac{1}{x}]}_{a}^{2 a}$
$= k [\frac{1}{2 a} - \frac{1}{a}]$
$W = - \frac{k}{2 a}$