What is the work-energy theorem for a block sliding on a frictionless incline?

AI Thread Summary
The discussion centers on applying the work-energy theorem to a block being pulled on a horizontal surface and then up a frictionless incline. The participants clarify the calculations for work done by tension, friction, and gravity, emphasizing the importance of understanding the signs and components of forces involved. A key point raised is that the total work done equals the change in kinetic energy, which must be negative when bringing the block to rest. Confusion arises over the correct values of kinetic energy at different points, with participants confirming that the kinetic energy before the incline is 200 J. Overall, the conversation highlights the challenges of mastering physics concepts and the significance of clear understanding in problem-solving.
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Homework Statement


A mass m = 16 kg is pulled along a horizontal floor, with a coefficient of kinetic friction μk = 0.09, for a distance d = 7.6 m. Then the mass is continued to be pulled up a frictionless incline that makes an angle θ = 34° with the horizontal. The entire time the massless rope used to pull the block is pulled parallel to the incline at an angle of θ = 34° (thus on the incline it is parallel to the surface) and has a tension T = 46 N.

blockonincline.png


1)
What is the work done by tension before the block gets to the incline?
2)
What is the work done by friction as the block slides on the flat horizontal surface?
3)
What is the speed of the block right before it begins to travel up the incline?
4)
How far up the incline does the block travel before coming to rest?
5)
What is the work done by gravity as it comes to rest?

Homework Equations

The Attempt at a Solution


First and foremost, I don't need help with every question. I've already solved 1-3. My problem is that I feel I'm just doing a 'trial and error' approach. I tend to get most of these questions wrong the first time and even after getting them right I don't usually feel that I've learned anything. For example, I just did a set of problems nearly identical to this one except that the horizontal plane was frictionless. When I look back on how I solved for the distance the block traveled up the incline, I don't even know why I solved it that way or why it even works.

Specifically, I found the kinetic energy of the block right before it hit the ramp, 478 J, and used the following equation: 478 + Wt + Wg +Wf = 0
Where Wt = work done by tension
Wg = work done by gravity
Wf = work done by friction

I don't even know why I used this equation. On the very next problem (the one above) I found myself wondering why the kinetic energy (200 J) was on the left side and questioning which side it should be on and whether it should be negative or positive. In short, I feel like I have absolutely no idea what's going on and I don't know how to fix this problem.
 
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Did you get KE = 478 or 200 at the bottom of the incline? I get 200.

The net work will equal the change in kinetic energy. Are we assuming the block starts from rest?

You’ve got work due to tension, and negative work due to friction. For the work due to tension, only consider the component in the direction of motion (W = T d cos34).

For friction be careful of your normal force. It’s not simply weight. You have to subtract the vertical component of tension.
 
Apparently I can't read. I didn't see that the incline plane is frictionless. So I got the answer for Q4, 4.79 meters.

Still, doing physics homework is a borderline nightmare for me. I feel utterly lost most of the time, and then I make it a thousand times harder on myself by making a simple mistake like this. I did the same thing last night where I had the sign of a force wrong and spent an hour losing my mind before finding the error.

tony873004 said:
Did you get KE = 478 or 200 at the bottom of the incline? I get 200.

It's 200J. 478 was for a previous problem where the flat plane is frictionless and the ramp has friction.

Let me ask you this: When bringing an object to a stop, does the total work just equal the initial kinetic energy, or is it the negative of the initial kinetic energy?
For example, in this problem if I set Wtotal equal to -200 then:
-200 = Wt - Wg
-200 = FtX - FgX
-200 = 46X - sin34(16)(9.81)X
-200 = X(46-87.77)
X = -200/-41.77
X = 4.79

So the work done has to be negative because it is done against the direction of motion, right? This didn't even click for me until just now.
 
Total work equals the change in KE. So if you start with +200 J of KE, and you end with 0, then your change in KE = -200, so your work must be -200 as well.
This is almost exactly like the example in your signature line. If you've got $200 in your bank account and you want to end with $0 in your account, you need to spend (negative, or red to an accountant) $200.
 
Are you studying the conservation of energy? What components constitute the energy sum?
 
tony873004 said:
Total work equals the change in KE. So if you start with +200 J of KE, and you end with 0, then your change in KE = -200, so your work must be -200 as well.

Okay, that makes sense. Thanks Tony.

Cumberland said:
Are you studying the conservation of energy?

Yes.

What components constitute the energy sum?

Are you referring to U+K?
 
Yes. What are the kinetic and potential energies when the block comes to rest? What is the only thing you don't know about the block at this point?

Edit: I just saw that you solved part 4 above. Sorry.
 
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