Question

A force $F=(10+0.5x)N$ acts on a particle in $X$ direction, where $x$ is in meters. Find the work done by this force during a displacement from $x=0$ to $x=2$.

Answer 1

Force $F=10+0.5x$

Work done $W=\int Fdx$

$W={\int }_0^2(10+0.5x)dx$

$W=[(10x+\frac{0.5x^2}{2}){]}_0^2$

So, $W=10\times (2-0)+\frac{0.5}{2}(2^2-0)=21J$