Calculating Kinetic Energy on an Inclined Ramp with Friction

[Dark Visitor]

Preston pushes a wheelbarrow weighing 500 N to the top of a 50 m ramp, inclined 20° with the horizontal, and leaves it. Tamara accidentally bumps the wheelbarrow. It slides back down the ramp, during which an 80 N frictional force acts on it over the 50 m. What is the wheelbarrow's kinetic energy at the bottom of the ramp? (g = 9.8 m/s2)

* 4550 J
* 6550 J
* 8150 J
* 13100 J


I am really confused with this one. I need help. Please get back to me a.s.a.p. with some explanations or anything that will be helpful. I don't even know where to start. Thanks.

 

[PhanthomJay]

Preston pushes a wheelbarrow weighing 500 N to the top of a 50 m ramp, inclined 20° with the horizontal, and leaves it. Tamara accidentally bumps the wheelbarrow. It slides back down the ramp, during which an 80 N frictional force acts on it over the 50 m. What is the wheelbarrow's kinetic energy at the bottom of the ramp? (g = 9.8 m/s2)

* 4550 J
* 6550 J
* 8150 J
* 13100 J


I am really confused with this one. I need help. Please get back to me a.s.a.p. with some explanations or anything that will be helpful. I don't even know where to start. Thanks.

Use the conservation of total energy equation, and watch your plus and minus signs.

 

[Dark Visitor]

What is that equation? I have conservation of energy equations, but I don't know which one to use, and you mentioned total energy.

 

[ladiesman217]

Yeah like PhanthomJay said, calculate the potential energy of the block when it is at the top of the ramp. (remember that the ramp is 50m long and inclined 20 degrees so the actual height is not 50! use sin function to find the height.)
then to find the resultant energy at the bottom you need to calculate how much energy is dissipated by the friction force.( Work done by friction )
just do step by step and you'll get it right :D

 

[Dark Visitor]

Yeah, I got the height like you said, using the sin function. But I still don't understand which Conservation of Energy equation to use, or how exactly to use it.

 

[PhanthomJay]

Yeah, I got the height like you said, using the sin function. But I still don't understand which Conservation of Energy equation to use, or how exactly to use it.

Which ones you got??

 

[Dark Visitor]

Total Mechanical Energy:
E = KE + U_mg


Law of Conservation of Energy for an Isolated System:
E (constant) = K + U_g + U_s

Law of Conservation of Energy including Energy transfers:
W + Q = \DeltaK + \DeltaU_g + \DeltaU_s

 

[PhanthomJay]

Law of Conservation of Energy for an Isolated System:
E (constant) = K + U_g + U_s

This is the Conservation of Energy theorem to be used when work is done by conservative forces only (like spring and gravity forces). It is often called the conservation of mechanical energy equation, where K + U is mechanical energy. Since you have a non conservative force acting (friction) that does work, don't use this one.

Law of Conservation of Energy including Energy transfers:
W + Q = \DeltaK + \DeltaU_g + \DeltaU_s

This is the more generalized form to be used when work is done by non-conservative forces (like friction). In your particular case, forget about Q (no heat added to the system), and there are no springs, and W (actually, it's W_{nc}, the work done by non conservative forces) is the work done by friction. Again, mind your plus and minus signs.

 

[Dark Visitor]

So are you saying that for this part, all I am focusing on is the work done by friction?

W_nc = W_f

 

[Dark Visitor]

Well, trying it out, I got:

W_{by F} = F*\Deltax*sin\theta
= (80 N)((50 m)(sin20))
= 1368.0806

Is that right? And if so, what do I need to find next?

 

[PhanthomJay]

So are you saying that for this part, all I am focusing on is the work done by friction?

W_nc = W_f

Yes,
W_f = \Delta K + \Delta U, or, if you prefer, K_{initial} + U_{initial} + W_f = K_{final} + U_{final} . I should have pointed out more clearly that the W in your earlier posted equation was wrong, it should have been W_{nc}.

 

[Dark Visitor]

So is what I just posted correct? And how do I find each individual piece to your equation?

 

[PhanthomJay]

Well, trying it out, I got:

W_{by F} = F*\Deltax*sin\theta
= (80 N)((50 m)(sin20))
= 1368.0806

Is that right?

No. The work done by friction is force times displacement times the angle between the force and displacemnt vectors (W = Fd cos theta). That angle theta is is not 20 degrees, since they are along the same incline. PLEASE watch your plus and minus signs, thanks.

 

[Dark Visitor]

Then what is the angle? That confuses me.

 

[PhanthomJay]

So is what I just posted correct?

see my later post, make correction

And how do I find each individual piece to your equation?

plug and chug, what's the equation for K, what's the equation for U?

 

[Dark Visitor]

see my later post, make correction plug and chug, what's the equation for K, what's the equation for U?

K = 1/2(m)(v)²

U_mg = mgy
(y is vertical height above/below y=0 position)

 

[PhanthomJay]

Then what is the angle? That confuses me.

Your posting faster than I can type or think. Which way does the wheel barrow go, down the slope or up the slope? Which way does the friction act, down the slope or up the slope? If friction is in the same direction as the displacement of the wheelbarrow, then theta is 0; if friction is opposite to the wheel barrow's motion, then theta is 180 degrees. So is the work plus or minus, plus or minus?

 

[Dark Visitor]

Friction should oppose the motion of the wheelbarrow, making it negative.

I also worked out the numbers, which I will show:

Ki = 0

Ui = (500 N)(17.101) = 8550.5036

Uf = (500 N)(0) = 0Now doesn't that mean that Kf = Ui + Wf?

 

[PhanthomJay]

Friction should oppose the motion of the wheelbarrow, making it negative.

I also worked out the numbers, which I will show:

Ki = 0

Ui = (500 N)(17.101) = 8550.5036

Uf = (500 N)(0) = 0


Now doesn't that mean that Kf = Ui + Wf?

Yes, correct. So now all you need to calculate is Wf, which is a negative term, Wf = work done by friction = (force of friction)*(distance through which that force acts) = ?

 

[Dark Visitor]

I can't remember exactly how I did it (it was due today, so I did what I could last night and turned it in.) I think I found W_f to be -4000, which my final answer turned out to be 4550 J, which was an answer, so I left it at that. Thanks for your help though. You got me that far.

 

[PhanthomJay]

I can't remember exactly how I did it (it was due today, so I did what I could last night and turned it in.) I think I found W_f to be -4000, which my final answer turned out to be 4550 J, which was an answer, so I left it at that. Thanks for your help though. You got me that far.

You're welcome, you got the right answer!

 

[Dark Visitor]

SERIOUSLY! SWEET! That just made my night! Thanks a lot man! I PASSED PHYSICS for the semester!