Question

At t=0, an object of mass m is at rest at x=0 on a horizontal, frictionless surface.

Answer 1

Answer:

a)$v=-T\dfrac{F_0}{m}(e^{-\dfrac{t}{T}} -1)$

b)$v=T\dfrac{F_0}{m}$

Explanation:

Given that

$F_x=F_0e^{-\dfrac{t}{T}}$

We know that force F given as

F=  m a

a=Acceleration

m=mass

$F=m\dfrac{dv}{dt}$

$F_0e^{-\dfrac{t}{T}}=m\dfrac{dv}{dt}$

$\dfrac{dv}{dt}=\dfrac{F_0}{m}e^{-\dfrac{t}{T}}$

$dv=\dfrac{F_0}{m}e^{-\dfrac{t}{T}}dt$

By applying boundary conditions

$\int_{0}^{v} dv=\int_{0}^{t}\dfrac{F_0}{m}e^{-\dfrac{t}{T}}dt$

$v=-T\left [\dfrac{F_0}{m}e^{-\dfrac{t}{T}} \right ]_0^{t}\\v=-T\dfrac{F_0}{m}(e^{-\dfrac{t}{T}} -1)$

The expression of velocity will be

$v=-T\dfrac{F_0}{m}(e^{-\dfrac{t}{T}} -1)$

When the time become so long

$e^{-\dfrac{t}{T}}=e^{-\dfrac{\infty }{T}}=0$

Therefore the velocity

$v=-T\dfrac{F_0}{m}(0 -1)$

$v=T\dfrac{F_0}{m}$