Question

A position dependent force $F=(x^2-3)\text{N}$ acts on a small body of mass $2\text{kg}$ and displaces it from $x=0$ to $x=5\text{m}$. The work done is

• A. $110\text{J}$
• B. $\frac{80}{3}\text{J}$
• C. $\frac{95}{2}\text{J}$
• D. Zero

Answer 1

The correct option is B $\frac{80}{3}\text{J}$

Work done by variable force is
$W={\int }_{x_1}^{x_2}FdxW={\int }_0^5(x^2-3)dx={[\frac{x^3}{3}-3x]}_0^5=\frac{125}{3}-15W=\frac{80}{3}\text{J}$
$\therefore \left(b\right)$

Why this question? "The work done by a variable force is given by $\int \overrightarrow{F}.d\overrightarrow{r}$ and in special case when both $\overrightarrow{F}$ and $\overrightarrow{r}$ are in $x$ direction, it is given by $W=\int Fdx$ "