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Piano sliding down an incline - work problem

  1. Oct 15, 2005 #1
    Hi, could someone please help me? Here is the problem:

    "A 265 kg piano slides 4.6 m down a 30° incline and is kept from accelerating by a man who is pushing back on it parallel to the incline (Fig. 6-36). The effective coefficient of kinetic friction is 0.40.

    (a) Calculate the force exerted by the man.
    (b) Calculate the work done by the man on the piano.
    (c) Calculate the work done by the friction force.
    (d) What is the work done by the force of gravity?
    (e) What is the net work done on the piano?

    I already found the answer to A, which is 399 N. The friction force was 900 N, btw (need that in C I think).
    Anyway, I'm trying to do B. Wouldn't the work just be zero, since the man's force has not caused any displacement of the piano. W = F d cos (angle). so with the d = 0, W would be zero....but I answered that in my Web Assign and it says that is wrong!

    The 4.6 m that the piano slides down is because of the force of gravity, so I understand that would have work, but I don't get why the man would, since he's not displacing the piano, he's using his force to STOP the piano from moving. Could someone explain B to me?? thank you so much for any help
     
    Last edited: Oct 15, 2005
  2. jcsd
  3. Oct 15, 2005 #2
    Remember that the piano is kept from accelerating not from moving - the man is making sure the piano moves with a constant velocity.
     
  4. Oct 16, 2005 #3
    Hm, I see..I did not read closely enough. Doh

    so wouldnt W = f d cos (angle) be W = 399 X 4.6 X cos 30 ?
    I got 1589.5, 1590 with 3 sig figs...but web assign says that is wrong. what am I doing wrong?
     
  5. Oct 16, 2005 #4
    think about it again..... what's the formula of work????
     
  6. Oct 16, 2005 #5
    W = Fd when F is parallel to displacement

    or W = F d cos (angle)

    Im so confused. The force should be parallel to the displacement...so I tried 4.6 X 399 = 1840 J, and web assign says that is wrong too! I have one more guess before I don't get any more tries, lol. I don't know what i"m doing wrong.
     
  7. Oct 16, 2005 #6
    the equation that you're supposed to use is

    W = F d

    Didn't you get the distance in the problem? Why did you try to find it with trig.? If you have to find d with cosine, then what's 4.6 m for?????
     
    Last edited: Oct 16, 2005
  8. Oct 16, 2005 #7
    Yea, I was confused because the book says W = fdcos(angle) but that's only when the force is not parallel to the displacement, I think.

    I tried it W = F d or 399 X 4.6 and that was wrong too.
     
  9. Oct 16, 2005 #8
    is the answer 1834.802 ??????
     
  10. Oct 16, 2005 #9
    I don't know. I got 1835.4 when I multiplied F X d. How did you get 1834.8?
     
  11. Oct 16, 2005 #10
    I saw that you round up the decimal. I did it without rounding anything.
     
  12. Oct 16, 2005 #11
    Oh crap...I just figured out what I needed.

    A simple negative sign!!!!

    The work is in done in the force opposite of the displacement so it is negative. Finally got it right. All that headache for a damn - sign, lol
     
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