Piano sliding down an incline - work problem

AI Thread Summary
A 265 kg piano slides down a 30° incline, with a man pushing against it to prevent acceleration. The man exerts a force of 399 N, while the friction force is 900 N. The confusion arises in calculating the work done by the man, as he believes it should be zero since he is not displacing the piano. However, the correct approach involves recognizing that the work done is negative due to the force acting opposite to the displacement, leading to a final calculation of approximately -1835 J. Understanding that work is negative when opposing displacement clarified the problem for the participant.
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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
 
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Remember that the piano is kept from accelerating not from moving - the man is making sure the piano moves with a constant velocity.
 
Hm, I see..I did not read closely enough. Doh

so wouldn't 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?
 
think about it again... what's the formula of work?
 
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.
 
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?
 
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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.
 
is the answer 1834.802 ?
 
I don't know. I got 1835.4 when I multiplied F X d. How did you get 1834.8?
 
  • #10
I saw that you round up the decimal. I did it without rounding anything.
 
  • #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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