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Validation of a Work/Energy Problem

  • Thread starter Mattchu
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  • #1
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Hi, this is my first post here. I was just wondering if anyone could validate my work done on this problem. (I have a feeling I missed something)

Here is the problem: A box of mass m is released from rest at Point A, the top of a long, frictionless slide. Point A is at height H above the level of Points B and C. Although the slide is frictionless, the horizontal surface from Point B to C is not. The coeffiecent of kinetic friction between the box and this surface is [mu] sub k, and the horizontal distance between Points B and C is x.

A) Find the speed of the box when its height above Point B is 1/2H .

B) Find the speed of the box when it reaches Point B.

[Edit in:]
C) Determine the value of [mu]k so that the box comes to rest at Point C.
[/Edit]

D) Now assume that Points B and C were not on the same horizontal level. In particular, assume that the surface from B to C had a uniform upward slope so that Point C was still at a horizontal distance of x from B but not at a vertical height of y above B. Answer the question posed in part (c).

E) If the slide were not frictionless, determine the work done by friction as the box moved from Point A to Point B if the speed of the box as it reached Point B was half the speed calculated in part (b).

My answers in variable form (I'm guessing that's how my teacher wanted them answered):

A) PE sub a + KE sub a + W = PE sub b + KE sub b
mg1/2h sub a = 1/2mv^2
[squ] = square root, not check
[squ] 2(g1/2h sub a) = v

B) gh sub a = 1/2v^2
[squ]2(gh sub a) = v

I'll just post my answer for the last two here because posting all of the work in order would take too long...

C) [mu] sub k = -(v sub c ^2 - v sub b^2 / N)

D) [mu] sub k = -(gh + 1/2v sub c ^2 - 1/2v sub b^2 / N)


[Edit in:]
E) fk = -(mvb^2 / 4mgha)
[/Edit]


I believe my problem is I'm not counting the force w sin [the] in the conservation of energy stuff but I'm not sure.


I'd appreciate any help. Thanks in advance.
 
Last edited:

Answers and Replies

  • #2
NateTG
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You can use the [ sub ] and [ /sub ] tags to create subscript.

Part B:

You need to account for the work done by friction as the box is sliding from B to C.

Part C:

You need to include not only the extra energy to raise the box, but also the extra friction from the increased normal force (if any)

Part D:

You should ignore part C for this answer. It's effectively asking how much energy would be lost if the speed was halved.
 
  • #3
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Part B: The slide from Points A to B was frictionless, though.

Part C: When you say energy to raise the box, you mean energy of potential? If so, I do have that in the equation (gh). I eliminated m from both sides.

Part D: Actually I did ignore part C, but I ended up with similar looking answers.
 
  • #4
Doc Al
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I found your original statement of the problem somewhat puzzling, since you mention a horizontal surface from B to C, but then it never comes up.
Originally posted by Mattchu
Part B: The slide from Points A to B was frictionless, though.
You solved that fine; VB= √(2gH)

Part C: When you say energy to raise the box, you mean energy of potential? If so, I do have that in the equation (gh). I eliminated m from both sides.
I do not understand the problem. Can you restate what is being asked for?
Part D: Actually I did ignore part C, but I ended up with similar looking answers.
Reread NateTG's response. (Point C is not involved.)
 
  • #5
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Thanks for bringing that up, Doc Al, because I reread my questions and answers and found that I had not posted question c or answer e. Now they should all be in the right places. Hopefully this clarifies any confusion you had with the problem.

I edited my first post if you want to take a look at it again.
 
  • #6
Doc Al
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Now that's better.

C) Here's how I'd do it. All the kinetic energy at B gets dissipated as work done by friction:
KEB = FfrictionX
mgH = μmgX
μ = H/X

D)I assume you meant to say that the position of point C is X (horizontal) and Y (vertical) from point B? In this case, the final energy at C = mgy (potential); the distance along which the friction acts is D = √(X2 + Y2); and the normal force is mg cosθ = mg(X/D). Thus:

mgH-mgY= μmg(X/D)D = μmgX
μ = (H-Y)/X (interesting, eh?)

E) if the speed is cut in half, what happens to the KE at the bottom compared to the no friction case? That loss in KE equals the work done by friction.
 
  • #7
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Two things I don't understand from your post: why did you substitute mgH for the value of KEb in question c? I assume you meant 1/2mvb^2.

Also, on question d, why did you put the force as mg cos [the]? Isn't it mg sin [the] when something is moving up an inclined plane?
 
  • #8
Doc Al
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Originally posted by Mattchu
Two things I don't understand from your post: why did you substitute mgH for the value of KEb in question c? I assume you meant 1/2mvb^2.
I substituted mgH because that's what the KE equals! You have to solve the problem in terms of what is given, which is the height H. (Leaving KE = 1/2mvb^2 is true by definition, but does not solve the problem. It's kind of lazy! )
Also, on question d, why did you put the force as mg cos [the]? Isn't it mg sin [the] when something is moving up an inclined plane?
What is needed is the normal force, which is perpendicular to the plane.
 

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