Question

A particle moving along the x-axis is acted upon by a single force $F=F_0{e}^{-kx}$, where $F_{0}andk$ are constants. The particle is released from rest at x= 0. It will attain a maximum kinetic energy o:

• A. $F_0/k$
• B. $F_0/e^2$
• C. $k{F}_0$
• D. $\frac{1}{2}(k{F}_0{)}^2$

Answer 1

The correct option is A $F_0/k$

Newton second low

$F=F_0{e}^{-kx}$

Kinetic energy $K.E=\frac{1}{2}m{v}^2$

$F=m\frac{dv}{dt}=m\frac{dv}{dx}\frac{dx}{dt}[\frac{dx}{dt}=v]$

$F.e^{-kx}=mv\frac{dv}{dx}$

$F_0{e}^{-kx}dx=mvdv$

$\frac{F_0}{k}{e}^{-kx}=\frac{m{v}^2}{2}$

$K.E=\left(\frac{F_0}{k}\right)e^{-kx}$

$K.E$ is maximum $\left(\frac{F_0}{k}\right)$