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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?

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?

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