Force and work homework problem

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SUMMARY

The discussion focuses on calculating the force exerted by a man pushing a 330kg piano down a 28-degree incline, where the effective coefficient of kinetic friction is 0.40. The user initially calculated the force using the formula F = mgsin(theta) and obtained a value of 3237.3N. However, the user questioned the correctness of this approach and the subsequent calculations for work done by the man and the friction force. The final calculation for work done by friction resulted in -2472.8J, which the user found questionable.

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  • Understanding of Newton's laws of motion
  • Familiarity with trigonometric functions in physics
  • Knowledge of work-energy principles
  • Basic understanding of friction and its coefficients
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  • Review the calculations for force using F = mgsin(theta) with correct angle adjustments
  • Study the work-energy theorem and its application in inclined plane problems
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Students studying physics, particularly those focusing on mechanics, as well as educators seeking to clarify concepts related to forces, work, and friction on inclined planes.

Farm4Life
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I have all of my question pretty much divided out into formulas and just need some help with with what i am putting into it.

The question reads, a 330kg piano slides 3.6m down a 28degree incline and is kept from accelerating by a man who is pushing back on it parallel to the incline. The effective coefficient of kinetic friction is 0.40. Calculate a. the force exerted by the man.

Since the man is parallel to the incline, his direction would be the same as mgsin(theta) wouldn't it?

I got for the force, F=mgsin(theta)
=330kg(9.81m/s2)sin 90
=3237.3N
Is this correct? or should i be using something different to derive the force? I think this is right though.

b. The work done by the man on the piano.

Work=Fdcos(theta), but theta would be zero right because the angle created between the man and the normal is 90, and cos90= 0

So, work would be zero right?

c. the work done by the friction force.

w=-fd Would you use the force of the man for the equation?

N+Fy-mg=0
N=mg-Fsintheta
N=330kg(9.81m/s2)-3237.3N(sin28)
N=3237.3-1519.6
N=1717.33

f=0.40(1717.33)=686.9

W(friction)=-686.9N(3.6m)=-2472.8J
This answer seems a bit large, I am not sure if i did something wrong, or if that seems like it should be. What would be your advice please?
 
Last edited:
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There is a few more questions to the problem, i have worked though them, but aren't very useful if i can't get these values correct.
Thanks!
 

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