How is a Conservation of Energy Problem Solved?

[bodensee9]

problem solved. Thanks!

 

[G01]

At first glance I see one error in your energy equation.

In the potential term you need the height difference, which is:

(Lsin30) not (h + Lsin30)

 

[bodensee9]

Hello:

But point A is on the ground, even below the beginning of the friction part. So wouldn't the potential different be (h + Lsin30)?
Thanks.

 

[G01]

Its hard to tell what you mean since I can't see the image yet, but from your description the frictional part if the path is on the incline, correct?

If so to find the change in PE, you need the change in height caused by sliding from A to B. This is Lsin30. Draw a picture and see for yourself.

Of course, I could be thinking of some different set up to this problem. Its hard to tell without the image.

 

[bodensee9]

Hello:

Sorry, so the box is on the flat part at the beginning. Point A is there. Then, the incline is initially frictionless up to a vertical height h, and then the friction part begins. The friction part has length L. So if I were to start from the flat part, wouldn't the change in PE = mg(h + Lsin30)? (because the box has to climb up a height h first to reach the friction part). Thanks.

 

[G01]

Hello:

Sorry, so the box is on the flat part at the beginning. Point A is there. Then, the incline is initially frictionless up to a vertical height h, and then the friction part begins. The friction part has length L. So if I were to start from the flat part, wouldn't the change in PE = mg(h + Lsin30)? (because the box has to climb up a height h first to reach the friction part). Thanks.

OK right. Then you have the correct PE term. The answers should come out to be the same.

So, next step. What are the two different answers your getting? I'll have an easier time finding your mistake if I can see the answers that your getting.

 

[bodensee9]

Hello:

So, doing the old kinematics way, I got 3.5 m/s. But doing it the energy way, I have:
1/2*m*8^2 = m*9.8*(2 + .75sin30) + 1/2*m*v^2 + m*9.8*cos30*.75. which comes out to be around 2.17?? Thanks.

 

[JazzFusion]

Hello:

So, doing the old kinematics way, I got 3.5 m/s. But doing it the energy way, I have:
1/2*m*8^2 = m*9.8*(2 + .75sin30) + 1/2*m*v^2 + m*9.8*cos30*.75. which comes out to be around 2.17?? Thanks.

Your set-up of the problem is correct and your understanding of the physics is excellent. Good job!

Your problem is in the execution of your approach, in the section I bolded above. The work done by friction is
= F * \Delta r,

= \mu_k * n * \Delta r

= (0.4) * m (9.8 m/s2) * cos30 * (0.75 m)

You forgot to use the coeeficient of kinetic friction in your equation, to find the work done by the force of friction.

Try that, and see if you get the exact same answer both ways.

 

[bodensee9]

Oh thanks! That took care of the problem!