contestada

At t=0, an object of mass m is at rest at x=0 on a horizontal, frictionless surface.
Starting at t=0, a horizontal force Fx=F0e−t/T is exerted on the object.

A)

Find an expression for the object's velocity at an arbitrary later time t.

Express your answer in terms of the variables F0, m, T, and t.

vx =
f0e−tTxm


B)

What is the object's velocity after a very long time has elapsed?

Express your answer in terms of the variables F0, m, and T.

vx =

Respuesta :

Answer:

a)[tex]v=-T\dfrac{F_0}{m}(e^{-\dfrac{t}{T}} -1)[/tex]

b)[tex]v=T\dfrac{F_0}{m}[/tex]

Explanation:

Given that

[tex]F_x=F_0e^{-\dfrac{t}{T}}[/tex]

We know that force F given as

F=  m a

a=Acceleration

m=mass

[tex]F=m\dfrac{dv}{dt}[/tex]

[tex]F_0e^{-\dfrac{t}{T}}=m\dfrac{dv}{dt}[/tex]

[tex]\dfrac{dv}{dt}=\dfrac{F_0}{m}e^{-\dfrac{t}{T}}[/tex]

[tex]dv=\dfrac{F_0}{m}e^{-\dfrac{t}{T}}dt[/tex]

By applying boundary conditions

[tex]\int_{0}^{v} dv=\int_{0}^{t}\dfrac{F_0}{m}e^{-\dfrac{t}{T}}dt[/tex]

[tex]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)[/tex]

The expression of velocity will be

[tex]v=-T\dfrac{F_0}{m}(e^{-\dfrac{t}{T}} -1)[/tex]

When the time become so long

[tex]e^{-\dfrac{t}{T}}=e^{-\dfrac{\infty }{T}}=0[/tex]

Therefore the velocity

[tex]v=-T\dfrac{F_0}{m}(0 -1)[/tex]

[tex]v=T\dfrac{F_0}{m}[/tex]