Work/Energy Help: Calculating Force, Work & Energy Applied to a Piano

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megaforcetkd
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Homework Statement



A 300 kg piano slides 4.0 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.

Picture attached

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

Homework Equations



Fg = Fn
Fgx = Fgsin(30)
Fgy = Fgcos(30)

W = F(displacement)cos@


The Attempt at a Solution



Fg = mg = (300 kg)(9.8 m/s2) = 2940 N
Fn = 2940 N

Fgx = 2940sin(30) = 1470 N
Fgy = 2940cos(30) = 2546.11 N

Ff = uFn = (0.4)(2940 N) = 1176 N

Fx = Fgx - Ff
= 1470 - 1176
= 294 N

Can't get any of it =/
 

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megaforcetkd said:

Homework Equations



Fg = Fn
Gravity acts vertically, while the normal force acts perpendicular to the incline surface. They are not equal. (If the surface was horizontal, then they would be equal.)
Fgx = Fgsin(30)
Fgy = Fgcos(30)
Good. Which of those is equal to the normal force?

W = F(displacement)cos@


The Attempt at a Solution



Fg = mg = (300 kg)(9.8 m/s2) = 2940 N
Fn = 2940 N
The wrong value for the normal force is messing you up.
 
ahhh right thanks :P.

i still can't seem to get the work done by the man on the piano though..

I do

W = (451.56)(4.0)cos(30)
 
Last edited:
Gah.. Still can't get it.. I'm on my last attempt.. If I screw it up I can't input anymore..
 
In part A I got 451.56 N

EDIT: Actually I inputted 451.96 but there's a 1% margin for error. I got that part right.
 
megaforcetkd said:
ahhh right thanks :P.

i still can't seem to get the work done by the man on the piano though..

I do

W = (451.56)(4.0)cos(30)
The force he exerts is parallel to the incline, so the angle (between force and displacement) is 0 degrees not 30.
 
I'm still getting it wrong using cos(0). On my last try.
 
Are inputing the correct sign? Since the man pushes up while the piano moves down, the work he does will be negative.

Doc Al said:
The force he exerts is parallel to the incline, so the angle (between force and displacement) is 0 degrees not 30.
More accurately, I should have said that the angle between force and displacement is 180 degrees, not really zero. :redface: (Sorry if that threw you off.)

(I usually separate the calculation of the magnitude of the work and the figuring out of its sign.)
 
What equation did you use to get 451 N in part A?
 
newtophysics said:
What equation did you use to get 451 N in part A?
Since the piano is not accelerating, the net force parallel to the incline must be zero. There are three forces to consider.
 
Oh wait, I just had a calculation error, thanks though!