Question

A force varying with distance is given as $F=a{e}^{-bx}$ acts on a particle of mass $m$ moving in a straight line. Find the work done on the particle if it is displaced by a distance $d$ from the origin.

• A. $\frac{a}{b}(1-e^{-bd})$
• B. $\frac{b}{a}(1-e^{-bd})$
• C. $\frac{a}{b}\left(e^{-bd}\right)$
• D. $\frac{b}{a}\left(e^{-bd}\right)$

Answer 1

The correct option is A $\frac{a}{b}(1-e^{-bd})$

As the applied force varies with displacement, so its work is given as

$W=\underset{0}{\overset{d}{\int }}F\cdot dx$

$W=\underset{0}{\overset{d}{\int }}a{e}^{-bx}dx$

$W=-\frac{a}{b}[e^{-bx}{]}_0^d=\frac{a}{b}(1-e^{-bd})$

So, option $\left(a\right)$ is correct.

Key concept:- In case of variable force the total work done will be sum of infinitesimal work over infinitesimal displacements. $W_{A\to B}=\int dW=\underset{A}{\overset{B}{\int }}\overrightarrow{F}\cdot d\overrightarrow{r}=\underset{A}{\overset{B}{\int }}\left(F\cos \theta \right)dr$