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Validation of a Work/Energy Problem

  1. Nov 14, 2003 #1
    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.

    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)

    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: Nov 14, 2003
  2. jcsd
  3. Nov 14, 2003 #2


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    Homework Helper

    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.
  4. Nov 14, 2003 #3
    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.
  5. Nov 14, 2003 #4

    Doc Al

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    Staff: Mentor

    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.
    You solved that fine; VB= √(2gH)
    I do not understand the problem. Can you restate what is being asked for?
    Reread NateTG's response. (Point C is not involved.)
  6. Nov 14, 2003 #5
    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.
  7. Nov 14, 2003 #6

    Doc Al

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    Staff: Mentor

    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.
  8. Nov 16, 2003 #7
    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?
  9. Nov 16, 2003 #8

    Doc Al

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    Staff: Mentor

    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! )
    What is needed is the normal force, which is perpendicular to the plane.
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