Question

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 $x=d$.

• A.
• B. b Joules
• C. O Joules
• D. None of these

Answer 1

The correct option is A

Given force varies with displacement of particle as $F=a+bx$ along $x$-direction and its asked work done

when particle moves from $x=0$ to $x=\alpha$.

$W=\overrightarrow{F}.\overrightarrow{\Delta s}\Rightarrow dW=\overrightarrow{F}.\overrightarrow{ds}\Rightarrow {\int }_o^{w}dw={\int }_o^d\overrightarrow{F}.\overrightarrow{dx}\Rightarrow w={\int }_o^{d}Fdscos\theta$

Since force also along x -direction.

So angle between $F$ and $ds$ = $0^{\circ }$

$Cos0=1$

$\Rightarrow w=\underset{0}{\overset{d}{\int }}F.ds$

$w=\underset{0}{\overset{d}{\int }}(a+bx)dx$

$w=ad+\frac{b{d}^2}{2}$

$w=d(a+\frac{bd}{2})$