Dynamics Prob force is a function of time

[fKnight]

Homework Statement

A block is attached to a rope through a single pulley connected to a motor pulling with a force F(t), t in seconds, when the block is falling at a specific velocity > 0 the motors is applied to stop, find the time it takes for v=0.

neglect mass of pulley & rope.

Homework Equations

The Attempt at a Solution

Is this as simple as setting F(t)=m*g? or would I need to set the forces in the y direction = to mass * accel(in y) then take the integral of dv from v to 0 = the integral of a dt from 0 to t. then solve for t?

 

[cepheid]

The motor isn't turned on until the block is falling at some rate, which we'll call the initial velocity v₀. So, given that the block is initally falling at that speed, how long does it take the motor to decelerate it? Let the time the motor is first turned on be t = 0. Let the stopping time, which we are trying to solve for, be t_s. Well, the change in velocity in that time interval is given by the integral of the acceleration:

[tex] v(t_s) = v_0 + \int_0^{t_s} a(t)\, dt [/tex]​

Taking downward to be the positive direction, we know that a(t) = F_net/m, where m is the mass of the block, and F_net is the net force on it:

[tex] a(t) = \frac{F_{\textrm{net}}}{m} = \frac{mg - F(t)}{m}[/tex]​

Since the speed at the stopping time is zero, the integral equation becomes:

[tex]\Delta v = -v_0 = \int_0^{t_s} g - \frac{F(t)}{m} \, dt [/tex]​

Eventually you get:

[tex] v_0 + gt_s = \frac{1}{m}\int_0^{t_s} F(t)\, dt [/tex]​

To be honest, I have no idea what to do with that without knowing the functional form of F(t)

 

[fKnight]

Nevermind, I think I figured it out.