MCAT question on Work

[SilvrGrey330]

Confused on Work, been seeing so many different definations im getting confused, please someone clarify.

Im trying to get all the formulas for work correct, heres what i have so far:

W = Fnet * d * cos & (& is theta)

W(gravity) = none , because & = 90 degrees
W(Normal force) = none , again because & = 90 degress

W(friction) = - F(friction) * d (- because & is 180 degrees)

Work energy theorem = delta KE = 1/2 mVf^2 - 1/2 mVi^2

when exactly do we use the Work energy theorem???

W(spring) = 1/2 k(Xf^2 - Xi^2)


Im confused over W(gravity) being none because I also thought that W(gravity) is mgh which is potential energy, can someone clarify work or gravity and potential energy for me and explain WET as well, i know this is a lot of questions in one, but its very confusing to me. Thanks!

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[nishi]

work is the product of the component of the force that acts though the displacement of an object. if a force acts at an angle to an object, on the compenent of the force vector that parallel to the objects displacement is used in the f x d =W equation (this ends up being fcosangle=work).
for a box at rest on the floor, gravity acts down, the object doesn't move down, so ther is no displacement. that's why the work done by the force of gravity is zero.

energy is the capacity to do work. if an object is at a height h, gravity can do work on it if the object were to fall because gravity acts down and the object will also move down when it falls. an object on a floor has no PE because gravity can't make it go down any more. the displacement will always be zero.

the total work done on an object equals the change in kinetic energy.
if an object with vi (inital velocity), vf, a, and distance:
vf ^2 = vi^2 +2ad
distance would equal ( vf ^ 2 - vi ^ 2 ) (1/2) (1/a)
and force = ma
so a = F/m
substuting a gives

( vf ^ 2 - vi ^ 2 ) (1/2) (m/F) = distance
if you multiply both sides by F you will get work on the left side. on the right you will be left w/ KE change

this only applies if all the forces are conservative.

 

[SilvrGrey330]

Thats cool, thanks for showing that derivative.

So let me get this straight,

There CAN be work done by gravity "During" flight from a free-falling object, because its moving through a height h, so in that instance, for free-falling objects, the Work = -mgh (assuming up is positive and gravity is -) which is also considered a Change in PE?

But when talking about objects at rest on ground, for example Block on table, at rest, the Fnet = 0, and W of gravity is also Zero since the Theto between the block and the table is 90 degrees.

When talking about a block on an frictionless inclined plane, the Work of gravity would be the force causing the object to move done,
so W= (mg sin&)(cos &)d

is this all Correct? FOr some reason W=(mg sin&)(cos&)d doesnt sound write, can someone correct me if im wrong. Thanks again!

 

[Daichi Katase]

The last part is correct.
mg sin(theta)= Force
W = F*d cos (theta)

but theta in the second case is 0, because the force applied and motion are at a 0 degree angle.

so.....W = f*d

simple as that

 

[Shrike]

I think y'all have worked it out. Nish's and chubby's answers are correct, and should be helpful. If you want more, I posted a complete explanation (in my opinion) of how to deal with work energy on the MCAT, in response to the same question, at http://forums.studentdoctor.net/showthread.php?t=175282&page=3&pp=29.

Enjoy.

 

[SilvrGrey330]

cool thanks, ya im gonna hit up some practice problems and drill this concept, i dont wanna miss these points.