What is the work done by the force of gravity? PLEASE HELP!!!! physics.........just one question :)?

Question

A 250 kg piano slides 4.2 m down a 30° 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.

(a) Calculate the force exerted by the man.
= 376.295 N

(b) Calculate the work done by the man on the piano.
= -1580.439 J

(c) Calculate the work done by the friction force.
= -3564.561 J

(d) What is the work done by the force of gravity?
= ... J

(e) What is the net work done on the piano?
= ... J


I got the first 3 questions; however, how do i find the work done by the force of gravity???????????????
i am so confused . i don't get it so PLEASE HELP!!!!!
I can't solve for the last one, unless if i find the work done by the force of gavity...

Answer 1

To get the work done by ANY particular force (such as the force of gravity), do this:

1) Figure out the component of the force that is parallel to the direction of motion. (In this case, the component of the gravitational force that points down the slope.)

2) Multiply that component times the distance moved.

In this case, you want the component of the gravitational force which is parallel to the incline. This is always mg·sinθ, where θ is the angle of the incline.

So, the work done by gravity on the piano is:
F_parallel middot; d
= (mg·sinθ)(4.2 m)

The only other thing to remember is: if the parallel component points OPPOSITE of the direction of motion, then the work done by that force is NEGATIVE. In this case of gravity, the parallel component is always downhill. In this problem, that's the same direction that the piano's moving, so the work done (by gravity) is positive.

[EDIT -- Other responders are correct too: you can get the same answer by multiplying the weight times the vertical distance the piano descended (which is: (4.2m)(sinθ)). You can either take the component of force that is parallel to the distance; or the component of the distance that's parallel to the force. You get the same answer either way. ]

Answer 2

The work done by gravity is just the force times the displacement

The down-slope component of gravity is mg sin (elevation)
so work done by gravity = (mg sin (elevation)) * d

The force of friction is mu * normal force
= mu * mg cos (elevation)
so the work (negative) done is mu * dmg cos (elevation)

The net work will then come out to be zero if you did it right because the kinetic energy didn't change. I don't think you could have done it right, though, if you calculated the man's push before you calculated gravity. You need to use the fact that the total work is zero to figure out how much negative work the man did.

His work must be work gravity - work friction
= mgd (sin (elevation) - mu cos (elevation))

Another way to do the problem is to treat gravity as a potential field. Potential fields don't do work per say. Then the total work will come out to be negative--the amount of potential energy lost by the piano.

Answer 3

The work done by gravity is simply the mass of the piano multiplied by the vertical distance it descends. Since the piano is motionless both before and after this adventure, the net work done on it is zero.