Piano sliding down an incline - work problem

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Homework Help Overview

The problem involves a 265 kg piano sliding down a 30° incline, with a man exerting a force to prevent its acceleration. The effective coefficient of kinetic friction is given as 0.40. Participants are discussing how to calculate the work done by the man, the friction force, and the force of gravity, among other related aspects.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants are attempting to calculate the work done by the man and are questioning the implications of the force being applied while the piano is moving at constant velocity. There is confusion regarding the correct application of the work formula, particularly in relation to the angle and direction of forces.

Discussion Status

Some participants have provided guidance on the work formula and its application, while others are exploring different interpretations of the problem. There is a recognition of the need to account for the direction of the force relative to displacement, leading to discussions about negative work.

Contextual Notes

Participants are navigating the constraints of the problem, including the specific values for forces and distances provided, as well as the implications of the piano moving at constant velocity despite the applied force. There is an emphasis on understanding the relationship between force, displacement, and work in this context.

confusedbyphysics
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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
 
Last edited:
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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?
 
Last edited:
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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