# Calc related question before hw is due please

1. Oct 3, 2011

### Fjolvar

A block is projected up an incline with speed Vo, find the time it takes to return to its initial position. The angle of incline is θ and coefficient of friction μk.

So I figured this problem has to be done in two parts. I integrated through F=ma to find the time it takes from vo to v=o when the block comes to rest. Then I integrated to find the distance it traveled x to reach that point. I figured all I would have to do now is integrate over the distance x but in the other direction with the coefficient of friction now in the opposite direction since the block is moving back down to it's initial position..

So here is my problem. I wrote dx2/dt2 = -gsinθ+μkgcosθ and I have x=Vo2 / 2(gsinθ+μkgcosθ) for the distance travelled. How do I find time? I can't just integrate twice to get x from dx2/dt2.. My calculus knowledge is failing me at the moment. Any help would be greatly appreciated.

2. Oct 3, 2011

### Fjolvar

Any ideas? Perhaps I could integrate velocity first from 0 to Vf and then integrate over X?

3. Oct 3, 2011

### Ray Vickson

Look at the acceleration, d^2 x/dt^2. Is it a constant, independent of x?

Have you really though carefully about the exact form of the frictional force? Your form looks disastrously wrong to me.

RGV

4. Oct 3, 2011

### SammyS

Staff Emeritus
If this is for a physics class, then I suggest using kinematic equations.

If it's for a Calculus class, then I suggest doing the integrations.

As the block moves up the incline, friction and gravity both act together to slow the block so they both have the same sign. So acceleration is given by
dx2/dt2 = -g  sin(θ) - μk g  cos(θ) .​
x=V02 / {2(g  sinθ+μk g  cosθ)} should give the distance that the block moves up the incline.

On the way down the incline you have
dx2/dt2 = -g  sin(θ) + μk g  cos(θ) ,​
because the friction opposes gravity.

You should be able to use the same kinematic equation (with the new acceleration) to solve for velocity, v, as the block returns to its original position.

Then either integrate
dx2/dt2 = -g  sin(θ) + μk g  cos(θ)​
once to get
dx/dt = (-g  sin(θ) + μk g  cos(θ))t +C , where v = dx/dt ,​
or use the kinematic equation
v - v0 = a  t , where v0 = 0 .​
Solve for t.