Question

A force $\overrightarrow{F}=(cx-3.00x^2)\hat{i}$ acts on a particle as the particle moves along $x-$ axis, with $\overrightarrow{F}$ in Newtons, $x$ in meters and $c$ is a constant. At $x=0$, the particle's kinetic energy is $20.0\text{J}$ and at $x=3.00\text{m}$, it is $11.00\text{J}$. Then, the value of $c$ is

• A. $1\text{N/m}$
• B. $2\text{N/m}$
• C. $3\text{N/m}$
• D. $4\text{N/m}$

Answer 1

The correct option is D $4\text{N/m}$

Given, $\overrightarrow{F}=(cx-3.00x^2)\hat{i}$
Kinetic energy at $x=0\text{m}$ is $20\text{J}$
Kinetic energy at $x=3\text{m}$ is $11\text{J}$
When body moves from $x=0\text{m}$ to $x=3\text{m}$,
work done by the force is
$W={\int }_{x_1}^{x_2}F\left(x\right).dx$
$\Rightarrow W={\int }_0^3(cx-3x^2)dx$
$\Rightarrow W={(\frac{c{x}^2}{2}-\frac{3x^3}{3})}_0^3$
$\Rightarrow W=\frac{c}{2}(3{)}^2-(3{)}^3$
$\Rightarrow W=4.5c-27$
According to work-energy theorem,
work done = change in KE.
$\Rightarrow 4.5c-27=(11-20)$
$\Rightarrow 4.5c-27=-9$
$\Rightarrow c=\frac{-9+27}{4.5}$
$\Rightarrow c=4\text{N/m}$
Hence option (d) is correct.