A position dependent force $F = 7 - 2 x + 3 x^{2}$ newton (where x is in meter) on a small body of mass 2 Kg & displace it from $x = 0$ to $x = 5$ m. Calculate the work done.

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Solution

$Step 1: Work done$
Work done for small displacement can be defined as:
$d W = F d x$ $. . . . (1)$
$Step 2: Integrate the equation$
Total work done can be calculated by integrating the equation $(1)$ with limits $x = 0$ to $x = 5 m$
$\int d W = 5 \int 0 F d x$
$\int d W = 5 \int 0 (7 - 2 x + 3 x^{2}) d x$
$\Rightarrow W = {[7 x - \frac{2 x^{2}}{2} + \frac{3 x^{3}}{3}]}_{0}^{5}$
$\Rightarrow W = 35 - 25 + 125 = 135 J$
Hence, the work done is $135 J$

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