Question

a particle of mass m sits at rest at x = 0. At time t = 0 a force given by F = Fe^(-t/T) is applied in the +x direction; F and T are constants. When t = T the force is removed. At this instant when the force is removed, (a) what is the speed of the particle and (b) where is it?

Answer 1

Step-by-step explanation:

Given:
`F=m\ddot{x}=Fe^{-(t)/(T)}`

Solving for
`\ddot{x}`:

`\ddot{x}=(F)/(m)e^{-\sqrt{(F)/(m) } t}`

where:

`T=\sqrt{(m)/(F)}`

Integrating to get
`\dot{x}` with initial conditions
`\dot{x}(0)=0`:

`\dot{x}=\sqrt{(F)/(m)}-\sqrt{(F)/(m)} e^{-\sqrt{(F)/(m)} t}`

Integrating to get x with initial conditions x(0) = 0:

`x=-1+\sqrt{(F)/(m)} t+e^{-\sqrt{(F)/(m)}t}`

When t=T:

`x=-1+\sqrt{(F)/(m)}\sqrt{(m)/(F)}+e^{-\sqrt{(F)/(m)}\sqrt{(m)/(F)}}=(1)/(e)`

`\dot{x}=\sqrt{(F)/(m)}-\sqrt{(F)/(m)} e^{-\sqrt{(F)/(m)}\sqrt{(m)/(F)}}=\sqrt{(F)/(m)}(1-(1)/(e))`