Dynamics Prob force is a function of time

AI Thread Summary
To determine the time it takes for a block to stop when a motor applies a force F(t), the net force and acceleration must be analyzed. The equation of motion involves integrating acceleration over time, leading to the relationship between initial velocity, gravitational force, and the applied force. The stopping time can be expressed as v_0 + gt_s = (1/m)∫_0^{t_s} F(t) dt, where v_0 is the initial velocity and g is the acceleration due to gravity. The functional form of F(t) is crucial for solving the integral. Ultimately, understanding the dynamics of the system allows for calculating the time required to bring the block to a stop.
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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?
 
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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 ts. Well, the change in velocity in that time interval is given by the integral of the acceleration:

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

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

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

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

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

Eventually you get:

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

To be honest, I have no idea what to do with that without knowing the functional form of F(t)
 
Nevermind, I think I figured it out.
 

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