Need help with a specific (slope) problem in dynamics

[mathchimp]

Homework Statement

An object is placed on a slope with initial velocity v=0m/s. Angle of the slope is 30 degrees and the coefficient of friction (or friction factor, not sure how it's said in English) is given as u=0.1*(x/m) where x is the path traversed and m is the mass of the object.

After how much time will the object stop moving?

Homework Equations

So far, what I know is:

friction: F1=u*m*g*cos(angle)
opposite of friction: F2=m*g*sin(angle)
acceleration in this case: a=(F2-F1)/m
velocity in general: v=dx/dt
acceleration in general: a=dv/dt

The Attempt at a Solution

I first tried calculating the acceleration. Since coefficient of friction is unknown and given as 0.1*(x/m), I end up with:

a=4.9-(x/m)*0.85

I checked it four times, pretty sure by now it's correct.
I now wanted to see if I can get something from a=dv/dt.

Integrating dt=dv/a, knowing that the initial velocity is zero, I wound up with t=1/a which I already knew.

I now tried using v=dx/dt which gave me:

t = (1/a)*(dx/dt)
tdt = (1/a)dx

Integrating it, again knowing that the initial path traversed is 0, I got

t^2/2=(1/a)*x

It didn't look very useful. Plugging a into the equation and making things look "pretty" gave me:

(4.9*t^2)/2 - 1/(2*m) = 1

Again, can't find the time. Mass unknown.

Since it was my 4th attempt at solving this, not knowing what to do and ready to break the laws of physics, I decide to derivate the equation and end up with t=0.204 s, which is of course false.

I'm wondering if there's anyone out there who can help me with this. The only thing I know is the solution, which is 3.41 seconds. I've tried solving this multiple times, asked my colleagues for help and searching the internet. Can't find anything useful. The book isn't very helpful with this.

 

[tnich]

a=4.9-(x/m)*0.85

That looks a differential equation:
$\ddot x = \frac g 2 - \frac {\sqrt 3 g} 2 ux$

 

[mathchimp]

That looks a differential equation:
$\ddot x = \frac g 2 - \frac {\sqrt 3 g} 2 ux$

Would you mind to explain a bit? We don't (ever) use differential equations in my physics class. How does that help me find the solution?

 

[tnich]

Would you mind to explain a bit? We don't (ever) use differential equations in my physics class. How does that help me find the solution?

Would you mind to explain a bit? We don't (ever) use differential equations in my physics class. How does that help me find the solution?

If you are not using differential equations in this class, then there is something we are misinterpreting in the problem statement. Since you have translated it to English, I can't tell you what is wrong with it (not that I could tell if I saw it in the original language).

By the way, I need to correct the differential equation I gave before. It should be:
$\ddot x= \frac g 2 - 0.1 \frac {\sqrt 3 x\,g} {2m}$

Another difficulty here is that this equation includes the mass m. This is due to the definition of u. $u= 0.1\frac x m$ is a very strange way to state a coefficient of friction. The coefficient of friction would not normally be dependent on m. This looks more like a coefficient that converts normal force to acceleration due to friction.

 

[mathchimp]

If you are not using differential equations in this class, then there is something we are misinterpreting in the problem statement. Since you have translated it to English, I can't tell you what is wrong with it (not that I could tell if I saw it in the original language).

By the way, I need to correct the differential equation I gave before. It should be:
$\ddot x= \frac g 2 - 0.1 \frac {\sqrt 3 x\,g} {2m}$

Another difficulty here is that this equation includes the mass m. This is due to the definition of u. $u= 0.1\frac x m$ is a very strange way to state a coefficient of friction. The coefficient of friction would not normally be dependent on m. This looks more like a coefficient that converts normal force to acceleration due to friction.

It is the coefficient of friction here, I've checked. There are many problems like this in my textbook but it usually cancels out with something and gives me a reasonable solution. I believe they expect the same thing here.

The only thing I perhaps didn't mention is that the body never makes it to the bottom of the slope because of u, but that's it. No more information.

Thank you for replying though, I assume problems like these won't appear on my exam.

 

[tnich]

It is the coefficient of friction here, I've checked. There are many problems like this in my textbook but it usually cancels out with something and gives me a reasonable solution. I believe they expect the same thing here.

The only thing I perhaps didn't mention is that the body never makes it to the bottom of the slope because of u, but that's it. No more information.

Thank you for replying though, I assume problems like these won't appear on my exam.

OK. Let's assume then that the acceleration due to friction is $a_f = u\,m\,g$. Then our differential equation is
$\ddot x= \frac g 2 - 0.1 \frac {\sqrt 3 \,g} 2 x$

This says that the acceleration changes with the distance traveled. And if you think about it, the speed of the object as it goes down the ramp has to increase from zero and then decrease back to zero, so the acceleration has to change to make that happen.

If you have no experience with differential equations, then try this - suppose that $x(t) = a\, cos(bt +c) + d$, where a, b, c and d are constants that you will have to determine.

Now differentiate $x(t)$ with respect to t get velocity $\dot x(t)$. You know that $\dot x(0) = 0$, so you can solve that equation for c. Given c, you can solve x(0) = 0 for d. That will simplify the equation for $\dot x$.

Differentiate again to get the acceleration $\ddot x(t)$. You can get another equation for $\ddot x(t)$ by substituting $x(t) = a\, cos(bt +c) + d$ in $\ddot x= \frac g 2 - 0.1 \frac {\sqrt 3 \,g} 2 x$. Now you have two equations for $\ddot x$ in terms of t. You already know c and d. Now you need to find values of a and b that will make the two equations look the same. Once you have those constants, you can find the values of t for which $\dot x(t) = 0$.

I made several edits in this post to correct errors.