Mechanics: Slope Friction Problem

[MathDestructor]

Homework Statement

Need Help with Work Energy Theorem

Relevant Equations

Attached Below.

So far, I have this:

For Part ii) I know that:

mgh = xmgsin(theta), but I don't know how to go further

 

[jbriggs444]

For Part ii) I know that:

mgh = xmgsin(theta), but I don't know how to go further

You know how much potential energy the object started with. Presumably you realize that it started and ended at rest -- that is, with zero kinetic energy.

So you can easily calculate how much energy was expended over the course of the downward slide. The work done over a path is given by the length of the path times the average [over distance] force along the path.

Does that give you a start?

You can make a useful observation about average force, given the nice way that this force is defined.

 

[haruspex]

You know how much potential energy the object started with. Presumably you realize that it started and ended at rest -- that is, with zero kinetic energy.

So you can easily calculate how much energy was expended over the course of the downward slide. The work done over a path is given by the length of the path times the average [over distance] force along the path.

Does that give you a start?

You can make a useful observation about average force, given the nice way that this force is defined.

Yes, except that "average force" is defined as average over time. You mean "average over distance".

 

[jbriggs444]

Yes, except that "average force" is defined as average over time. You mean "average over distance".

Which is why I specified an average over distance.

My intuition is also screaming that due to symmetry, the two averages will match in this case.

 

[haruspex]

Which is why I specified an average over distance.

So you did - sorry.

 

[haruspex]

My intuition is also screaming that due to symmetry, the two averages will match in this case.

Wouldn't it be SHM (until the bottom)? The two averages differ for that.

OTOH, looks like the question can be answered using a symmetry argument without needing to mention averages or perform integrals.

 

[MathDestructor]

Is this correct? How do I take the integral in the next part?

 

[haruspex]

Is this correct? How do I take the integral in the next part?

For the first part, your two methods are effectively the same (and correct).
For the next part, I gave rather a large clue in post #6. Apart from the constant force of gravity, the force is proportional to what? Does that ring any bells?

 

[MathDestructor]

For the first part, your two methods are effectively the same (and correct).
For the next part, I gave rather a large clue in post #6. Apart from the constant force of gravity, the force is proportional to what? Does that ring any bells?

I don't understand sorry

 

[MathDestructor]

Is my equation for Q3 right? If so how do I go about solving it?

 

[haruspex]

Is my equation for Q3 right? If so how do I go about solving it?

The easy way is to recognise your first equation in post #7 as being (almost) a well known standard equation, with $\ddot x$ on one side and a term proportional to -x on the other.

Failing that, your v(x) = equation in the middle if the second image can easily be turned into time = an integral of a function of x wrt x.
The first step in solving the integral is to get rid of the x term (as opposed to the x squared term) by a simple change of variable. That will give you a constant term instead. Post what you get.
Then think about another substitution to get rid of the square root. I'll let you have a go at that before any further hints.

 

[MathDestructor]

The easy way is to recognise your first equation in post #7 as being (almost) a well known standard equation, with $\ddot x$ on one side and a term proportional to -x on the other.

Failing that, your v(x) = equation in the middle if the second image can easily be turned into time = an integral of a function of x wrt x.
The first step in solving the integral is to get rid of the x term (as opposed to the x squared term) by a simple change of variable. That will give you a constant term instead. Post what you get.
Then think about another substitution to get rid of the square root. I'll let you have a go at that before any further hints.

What change of variable are you referring to? Could you please start me off with the first step?

 

[MathDestructor]

I got to this, what's next?

 

[haruspex]

View attachment 256660

I got to this, what's next?

A simple shift in x (adding a constant) can eliminate the linear term, giving an expression of the form $A\sqrt{1-cx^2}$.
A multiplicative substitution gets rid of c.
Do you see what to do from there?

 

[MathDestructor]

Yeah i got it, thanks !

 

[haruspex]

Yeah i got it, thanks !

Good.
Note that in the original problem if you take x as the displacement from the middle then you get the SHM ODE straight away.