Calculating Time for Train to Stop Moving Up Inclined Mine

[Aggie]

An engine is used to pull a train of 2 cars out of a mine. The floor of themine slopes upward at an angle of 30 degrees. Each car has a mass of 10^4 kg and travels without friction. The engine can exert a max force of 1.5 * 10^5 N

if the engineer trttles back so that the force exerted by the engine on car A decreases at the constant rate of 3N per second, how long before the train stops moving up the track? Assume the original speed was 3 meters per second

Net Force = F - rate
I think

 

[H_man]

Hmm...

Draw a force diagram, try setting out the relevant equations and it should be straight forward.

There is no friction so its just a matter of when the cars forward force is balanced by gravity pulling it down-hill.

 

[apelling]

Net Force = F - rate

You have not considered the force down the plane due to the component of weight in that direction. Once you've taken this from the F-rate*time (you must multiply the rate by time so that force decreases with time) you have the full expression for net force.

Then use net Force=ma, rearrange to get a=... and integrate with respect to time to find an expression for velocity (add the 3m/s initial velocity on here). This should leave you with a quadratic in time which you solve to find when the velocity is zero. (The answer I got was over 4 hours)

This question is complicated by the fact that acceleration is not uniform but changes with time so we cannot apply the equations of uniform acceleration here.

When I first read this question I made the same mistake as H_man and thought we needed to find the time till the forces are balanced. But of course at this point the train still has forward velocity and must be decelerated back to zero by a negative force down the plane.