Work Energy Problem

[DJBruce]

Here is a question from my physic's problem set. I was wondering if my solution is correct.

 

Question: How much work will Madison have to do to push DJ up a 22 degree ramp that is 6m long. DJ has a mass of 77.2 kg and a coefficient of friction is .29. First assume that she pushes at a constant velocity then solve again for an acceleration of 1m/s/s.

 

My solution is that work is equal to the change in energy.

 

[math]w=\Delta E[/math]

 

At the beginning of the hill there is no kinetic energy and no gravitational potential energy. At the top of the hill all the energy is gravitational potential energy so the change in energy is:

 

[math]\Delta E=mg\Delta y[/math]

 

I am going to assume that the bottom of the hill represents a initial height of 0m. The height at the top of hill is equal to:

 

[math] height = side opposite= hypotenuse(Sin(\theta ) [/math]

[math] height= 6(Sin(22^{o})=2.248m[/math]

 

Then the work done is:

 

[math] w=(77.2)(9.8)(2.248)=1700.5 J[/math]

 

Since the height is the same for both situations the change is energy is the same in both, and thus the work done is the same in both situations.

 

I was just wondering if these calculations were correct.

[Mr Skeptic]

Your calculations are correct for a frictionless surface. You are only taking into account the potential energy, not heat nor change in kinetic energy (as would result from continuous acceleration).

[Cap'n Refsmmat]

I don't think that's right. That may represent the net work (and I have doubts about that), but friction is contributing negative work (since it acts to oppose the motion), so the work required from Madison is greater. Also, [math]w = \Delta KE[/math], not total mechanical energy.

 

I'd draw a free-body diagram and work from there.

[Mr Skeptic]

Also, [math]w = \Delta KE[/math], not total mechanical energy.

 

Eh? That ignores potential energy. Properly, [math]w = \int_{\vec{a}}^{\vec{b}} \vec{F} \cdot d\vec{r}[/math] or in non-calculus form, [math]w = F_x \cdot \Delta x[/math]. The work done can increase potential energy or kinetic energy (heat from friction being kinetic energy on the molecular level).

[Cap'n Refsmmat]

Eh. What I said was true for net work: the net work is zero if the kinetic energy does not change. As for the work of one particular person, saying [imath]w= \Delta E[/imath] works if you don't consider friction or anything else, because the gravitational potential change is equal to the work done by the gravitational force. Friction messes things up a bit though.

[DJBruce]

I had a feeling that my solution was incorrect. However, before I spent the time to do it the longer way, I figured I would see if the easy way was correct. Thanks for the help guys.

[Mr Skeptic]

I think you can separately calculate the work to change gravitational potential energy, the work to overcome friction, and the work to accelerate the mass, and add them as appropriate.

[DJBruce]

I think you can separately calculate the work to change gravitational potential energy, the work to overcome friction, and the work to accelerate the mass, and add them as appropriate.

 

That's what I am doing, and it works. As I said I was hoping the problem was meant to be an easy trick one.

 

Thanks again for the help.