Question

a particle moving along the x axis is acted upon by a single force F= Foe^(-kx) . where Fo and k are constants.
The particle is released from rest at r=0. It will attain a maximum kinetic energy of:​

Answer 1

A particle moving along the x-axis is acted upon by a single force, F = $F_0e^{(-kx)}$, where $F_0$ and k are constants.

workdone by the particle = change in kinetic energy,

⇒∫F dx = Kf - Ki

as particle is released from rest so, Ki = 0

then, Kf = $\int\limits^x_0{F_0e^{-kx}}\,dx$

= $F_0\left[\frac{-e^{-kx}}{k}\right]^x_0$

= $F_0\left[-\frac{e^{-kx}}{k}+\frac{1}{k}\right]$

now, for maximum kinetic energy, force should be applied maximum.

Force will be maximum only when F'(x) = 0, -kFoe^-kx = 0, x = ∞

then, Kf = maximum kinetic energy = $F_0\left[-\frac{0}{k}+\frac{1}{k}\right]$

= $\frac{F_0}{k}$ [ ans]

Answer 2

$\huge\bold\purple{Answer:-}$

A particle moving along the x-axis is acted upon by a single force, F = F_0e^{(-kx)}, where F_0 and k are constants.

workdone by the particle = change in kinetic energy,

⇒∫F dx = Kf - Ki

as particle is released from rest so, Ki = 0

then, Kf = \int\limits^x_0{F_0e^{-kx}}\,dx

= F_0\left[\frac{-e^{-kx}}{k}\right]^x_0

= F_0\left[-\frac{e^{-kx}}{k}+\frac{1}{k}\right]

now, for maximum kinetic energy, force should be applied maximum.

Force will be maximum only when F'(x) = 0, -kFoe^-kx = 0, x = ∞

then, Kf = maximum kinetic energy = F_0\left[-\frac{0}{k}+\frac{1}{k}\right]

= \frac{F_0}{k} [ ans]