Calculating Force and Work of Man Pushing Piano

[crisicola]

Hi, this my first time asking a question, so here it goes:
A 255 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. 6-35). The effective coefficient of kinetic friction is 0.40.

(a) Calculate the force exerted by the man.
So far, I have this answer, by subtracting the Fparallel(mg*sin(30)) by Ffriction(mu*mg*cos(30)). [383.8 N]

(b) Calculate the work done by the man on the piano.
Got this answer, too, by multiplying the answer above by 4.6m. [-1764.6 J]

It's the next parts I got wrong.
(c) Calculate the work done by the friction force.
I figured Ffriction = mu * m * g * cos(30) = 865.7 * d = 3982.12 J, but that doesn't make sense. [?J]

(d) What is the work done by the force of gravity?
9.8 m/s^2 * 255g * 4.6m = 11495.4, it's incorrect [?J]

(e) What is the net work done on the piano?
This one I had no idea how to do altogether. J

Please help me out, thank you in advance!

 

[Doc Al]

It's the next parts I got wrong.
(c) Calculate the work done by the friction force.
I figured Ffriction = mu * m * g * cos(30) = 865.7 * d = 3982.12 J, but that doesn't make sense. [?J]

This is OK except for the sign. Since the friction points up the ramp, while the piano slides down the ramp, the work done is negative.

(d) What is the work done by the force of gravity?
9.8 m/s^2 * 255g * 4.6m = 11495.4, it's incorrect [?J]

When calculating work, you must use the component of the force parallel to the displacement.

(e) What is the net work done on the piano?
This one I had no idea how to do altogether. J

What's the net force on the piano?

 

[quicknote]

Hi there! It's great to see you tackling a physics problem. Let's take a look at your calculations and see where you may have gone wrong.

(a) Your calculation for the force exerted by the man is correct. This is the force required to counteract the component of gravity acting down the incline.

(b) Again, your calculation for the work done by the man is correct. This is the energy expended by the man in pushing the piano up the incline.

(c) To calculate the work done by the friction force, we need to use the formula W = F*d*cosθ, where F is the friction force, d is the distance traveled, and θ is the angle between the force and the direction of motion. In this case, the friction force is given by Ffriction = μ*mg*cosθ, where μ is the coefficient of kinetic friction, m is the mass of the piano, and g is the acceleration due to gravity. So the work done by the friction force would be W = μ*mg*cosθ*d*cosθ = μ*mg*d*cos^2θ. Plugging in the values, we get W = (0.40)*(255 kg)*(9.8 m/s^2)*(4.6 m)*cos^2(30°) = 792.8 J. This is the energy dissipated due to friction.

(d) The work done by the force of gravity is given by W = Fgd = mgd*cosθ, where Fg is the force of gravity, m is the mass of the piano, d is the distance traveled, and θ is the angle between the force and the direction of motion. In this case, θ is 30°, so we get W = (255 kg)*(9.8 m/s^2)*(4.6 m)*cos(30°) = 11495.4 J. This is the energy gained by the piano due to gravity.

(e) The net work done on the piano is the sum of the work done by all the forces acting on it. In this case, we have the work done by the man (−1764.6 J), the work done by the friction force (792.8 J), and the work done by the force of gravity (11495.4 J). So the net work done on the piano would be Wnet = −1764.6 J + 792.8 J +