Question

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 =

Answer 1

To find the expression for the object's velocity, integrate the equation of motion using Newton's second law, F = ma. The object's velocity after a very long time has elapsed is zero.

Step-by-step explanation:

To find the expression for the object's velocity, we need to integrate the equation of motion. Let's assume the initial velocity is v0 = 0 m/s. Using Newton's second law, F = ma, we have F = kx and ma = kx. Rearranging the equation, we have m(dv/dt) = kx. Dividing both sides by m and rearranging, we get dv/dt = (k/m)x.

Now, since F = F0e^(-t/T), we can write F/m = F0/(m)e^(-t/T). Substituting this value for F/m in the previous equation, we have dv/dt = F0/(m)e^(-t/T)x. Integrating both sides of the equation with respect to time, we get v = -F0/(mT)e^(-t/T)x + C, where C is the integration constant.

When a very long time has elapsed, the exponential term e^(-t/T) approaches zero. Therefore, the velocity at that time will be v = C. Since initially the object is at rest at x = 0, the integration constant is also zero, which means C = 0. Hence, the object's velocity after a very long time has elapsed is zero.