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

In summary: So you have KE, PE, any other energy terms?In summary, a mass of 16 kg is pulled along a horizontal floor with a coefficient of kinetic friction of 0.09 for a distance of 7.6 m. It is then pulled up a frictionless incline at an angle of 34° with a tension of 46 N, while the massless rope used to pull the block is parallel to the incline. The work done by tension before the block reaches the incline is 0 J, and the work done by friction as the block slides on the flat horizontal surface is -64.9 J. The speed of the block right before it begins to travel up the incline is 4.79
  • #1
Drakkith
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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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  • #2
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.
 
  • #3
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.
 
  • #4
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.
 
  • #5
Are you studying the conservation of energy? What components constitute the energy sum?
 
  • #6
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?
 
  • #7
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.
 
Last edited:
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1. What is the concept of "Block Sliding up an Incline" in physics?

The concept of "Block Sliding up an Incline" refers to the study of the motion of a block that is placed on an inclined surface and experiences the force of gravity pulling it down the incline. This concept is often used in physics to understand the relationship between forces, mass, and motion.

2. How is the acceleration of the block affected by the angle of the incline?

The acceleration of the block is affected by the angle of the incline as the steeper the incline, the greater the force of gravity acting on the block. This results in a higher acceleration of the block down the incline. However, the component of gravity pulling the block down the incline decreases as the angle decreases, resulting in a lower acceleration.

3. What is the significance of the coefficient of friction in "Block Sliding up an Incline"?

The coefficient of friction represents the amount of resistance between the block and the incline. It determines the force required to keep the block in place or to make it move up or down the incline. A higher coefficient of friction means more resistance and a lower coefficient of friction means less resistance.

4. How does the mass of the block affect its motion on the incline?

The mass of the block affects its motion on the incline as it determines the amount of force needed to accelerate the block. A heavier block requires more force to move up the incline, while a lighter block requires less force. However, the mass of the block does not affect the acceleration due to gravity.

5. Is there a specific formula to calculate the acceleration of a block sliding up an incline?

Yes, the formula to calculate the acceleration of a block sliding up an incline is a = g(sinθ - μcosθ), where a is the acceleration, g is the acceleration due to gravity (9.8 m/s²), θ is the angle of the incline, and μ is the coefficient of friction. This formula is based on Newton's Second Law of Motion, which states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass.

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