Motion up an Inclined Plane - Not sure if I did it right (1 Viewer)

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1. The problem statement, all variables and given/known data
A brick is projected up an inclined plane with an initial speed v0.

If the inclination of the plane is 30 degrees and the coefficient of sliding friction μ = 0.1, find the total time for the block to return to its original position.


2. Relevant equations



3. The attempt at a solution
Consider the x axis to be along the inclined plane. Then there are two forces acting along the x axis: the x component of gravity, and the friction.

X component of gravity: mgsin(30)
fricition: mgcos(30)μ, since mgcos(30) is the normal force on the brick

Then ma = -mgsin(30) - mgcos(30)μ or a = -gsin(30) - gcos(30)μ

a = dv/dt, so:

dv = (-gsin(30) - gcos(30)μ) dt

Integrating both sides gives

v = -gsin(30)t - gcos(30)μt + C, where C is a constant of integration

The initial conditions are at t = 0, v = v0 so C = V0

Then v = -4.905t - 0.85t + V0 = -5.755t + V0

Going up the incline, I set v = 0 and find that t = V0 / 5.755

I'm not sure if I can just double the above, because on the way up the incline gravity and friction are in the same direction. On the way down, they are in opposite directions.
 
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vela

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No, you can't double it for the reason you suspect. As it slides back down, it will cover the same distance, but its acceleration will be different, so it'll take more time getting back down.
 
So if I just to the same thing, but down the plane instead of up, then add the two times together, it should work.
 

vela

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It's a little different since you don't know the final velocity.
 
I didn't think of that...

I guess it would be better to find the distance the brick travels up the slope, then use that distance on the way back down.

So instead of using calculus to do this, on the way up I could find acceleration and distance.

Then on the way down, I could also find acceleration, and I'd have distance, so I could find time. I think, anyway.

How would this work if the force on the block wasn't constant though? The reason I wanted to use calculus is that in class we're dealing more with changing forces than static ones now.
 

vela

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In this problem, you have constant accelerations, so there's really no need to resort to calculus as you presumably have the kinematic equations for constant acceleration. What you did was essentially rederive one of those equations.

If the force isn't constant, you may need to use calculus to solve the equation. You'll see this when you study simple harmonic motion.
 

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