Calculating Energy Dissipated by Friction

[Sw0rDz]

Homework Statement

A 22.9-kg girl slides down a playground slide with a vertical drop of 3.29 m. When she reaches the bottom of the slide, her speed is 1.34 m/s. How much energy was dissipated by friction?

Homework Equations

I used \DeltaKE + \DeltaPE = 0

The Attempt at a Solution

I got KE_initial = 0
KE_final = 1/2 * 22.9kg * (1.34)² = 20.56

I got PE_inititial = m * g * h = 22.9 * 9.8 * 3.29 = 738.34
PE_final = 0

20.56 (Friction) = 738.34

Is that on the right track? Where would I go from there?

 

[gabbagabbahey]

Hi Sw0rDz, welcome to PF!

Homework Equations

I used \DeltaKE + \DeltaPE = 0

If you are using PE to represent gravitational potential energy, then there is another term that should be included in this equation; the work done by friction W_f:

\Delta\text{KE}+\Delta\text{PE}+W_f=0

make sense?

P.S. In the future, problem like this should go in the Introductory physics folder

 

[Doc Al]

Homework Equations

I used \DeltaKE + \DeltaPE = 0

This would be true if mechanical energy was conserved--there was no friction--but that's not the case here. How would you modify this to include energy dissipated by friction?

The Attempt at a Solution

I got KE_initial = 0
KE_final = 1/2 * 22.9kg * (1.34)² = 20.56

I got PE_inititial = m * g * h = 22.9 * 9.8 * 3.29 = 738.34
PE_final = 0

This looks good.

20.56 (Friction) = 738.34

I don't know where you got this or what it means.

Hint: Compare the initial mechanical energy with the final.

 

[Sw0rDz]

Hi Sw0rDz, welcome to PF!
If you are using PE to represent gravitational potential energy, then there is another term that should be included in this equation; the work done by friction W_f:

\Delta\text{KE}+\Delta\text{PE}+W_f=0

make sense?

P.S. In the future, problem like this should go in the Introductory physics folder

Thanks! That is what I needed.

Second part of the problem is:
If the slide is inclined at 20.5° with the horizontal, what is the coefficient of kinetic friction between the girl and the slide?

KE doesn't depend on the angle, so I left it the same.

I took PE to be m * g * cos(20.5) *h

Is that right?

 

[gabbagabbahey]

KE doesn't depend on the angle, so I left it the same.

I took PE to be m * g * cos(20.5) *h

Is that right?

No, neither energy term will depend on the angle...what will depend on the angle are the components of the gravitational and frictional forces parallel and normal to the slide.

 

[Sw0rDz]

Energy lost due to work will be the same. Just when trying to calculate the coefficient of friction will depend on the angle?

 

[gabbagabbahey]

What direction does the force of friction act in? What is the formula for finding its magnitude?

 

[Sw0rDz]

In my diagram, have it acting left (negative)

Previously I got.
W_friction = 717.79J
(Was right according to LON-CAPA and previous equation)

Now I set. 717.79 = \upsilon_KF * M * G * Cos(\Theta)

 

[gabbagabbahey]

Now I set. 717.79 = \upsilon_KF * M * G * Cos(\Theta)

Where did you get that formula from?

What is the general formula for frictional force? What is the general formula (involves an integral) for work done by a force?

 

[Sw0rDz]

F_friction = (Coefficient of KF) * (Force)
W = \int F * dl

Where dl = infinitive displacement.

 

[gabbagabbahey]

Right, and what is the normal force in this case?

 

[Sw0rDz]

F_n_o_r_m_a_l = M *G * Cos (\theta)

 

[gabbagabbahey]

You should probably use a lowercase 'g' so as not to confuse it with the universal gravitational constant, but yes.

So, F_f=\mu_f mg\cos\theta...what about its direction?

 

[Sw0rDz]

Negative because it acts against the positive kinetic energy.

Maybe..
<br /> -F_f=\mu_f mg\cos\theta<br />

 

[gabbagabbahey]

"Negative" isn't a direction, it's just a sign... does the frictional force act to the left? to the right? up? down? tangent to the slide? normal to the slide?

 

[Sw0rDz]

In my diagram. Positive direction is to the right. Negative is to the left. Friction goes to the left.
Tangent to the slide, I think..

 

[gabbagabbahey]

Directly to the left? Shouldn't it be going tangent to the slide, opposite to the girl's velocity?

 

[Sw0rDz]

Yes because friction is based of the Normal Force. And Normal force is perpendicular to the tangent line of the slide.

 

[gabbagabbahey]

Okay, so wha soes that tell you about \textbf{F}_f\cdot d\textbf{l}? And what do you get when you integrate it over the length of the slide? How long is the slide?

 

[Sw0rDz]

<br /> \int_0^{3.2} \left( ce_{KF} \cdot m\cdot g\cdot cos(\theta)\cdot h\right) dh<br /><br /> \left[ ce_{KF} \cdot m\cdotg\cdot g\cdot cos(\theta) \cdot \frac{1}{2} h^{2} \right]_{0}^{3.2}<br /> <br />

??

 

[gabbagabbahey]

3.29m is the height of the slide...you want to integrate along the path that the girl takes, which means you need to know the length of the slide (the hypoteneuse)...

 

[Sw0rDz]

Okay.
<br /> hyp. = sin\theta = \frac{hyp}{opp} = sin(20.5) = \frac{3.29}{x}<br />
<br /> x = \frac{sin(20.5)}{3.29}<br />

 

[gabbagabbahey]

You've got that backwards...

\sin\theta=\frac{\text{opp}}{\text{hyp}}\implies \text{hyp}=\frac{3.29\text{m}}{\sin\theta}[/itex]<br /> <br /> So, what expression do you get for W?...Compare that to your answer from part (a) and solve for \mu_f

 

[Sw0rDz]

Sorry. Thanks for the help. I wish I could figure this out and if not for the time, I probably would. Its roughly 2:20am and my brain has meant exhaustion. I'll check back at the thread after a good night sleep, even though I'll not be able to finish my homework on time.