Is the Work Done by Gravity Calculated Incorrectly?

[suspenc3]

Work Question..Anyone?!

Im doing a problem and I found the work done by an applied force to be 401J.
I then found the work done by gravity to be -330.8J
Does this mean that i made an error..Arent (Wg + Wa) suppose to = 0

 

[suspenc3]

There is a mass sitting on an incline and a force is applied so that it slides down at a constant speed.

I found F=267N
W=Fd...
W=401J

Wg=mgdcos(a)
Wg=-330N

Is d the total distance travelled..Or the y component of the distance travelled

 

[Pyrrhus]

Please post the problem statement!

 

[suspenc3]

Sure here is it

A 45kg block of ice slides down a frictionless incline of 1.5m long and 0.9m. A worker pushes up against the ice parallel to the incline, so that the block slides down at a constant speed.

 

[suspenc3]

a)find the magnitude of the workers force
b)how much work is done on the block by the workers force
c)" " the gravitational force on the block
d)the normal force on the block from the surface on the incline
e)the net force on the block

 

[Pyrrhus]

Ok so show me your work, and/or where did you get stuck?

 

[suspenc3]

drew the fbd's and came up with:

-mgsin(theta) - F = -ma (a=0)
-mgsin(theta) = F
plug in numbers

F = 267.5N
Wf = Fd
Wf = 267.5N(1.5m)
Wf = 401J

Wg = mgdcos(phi)
Wg = 45kg(9.8m/s)(1.5m)(cos120)
Wg = -330.8J

This can't be correct because Wg +Wf = 0?

 

[Pyrrhus]

i suppose the 1.5 and 0.9 are sides of the triangle right?

 

[suspenc3]

1.5 = hypotenuse
0.9 = height

 

[Pyrrhus]

Ok i see the problem the mgcos(phi) component of gravity does not do work!, because it is perpendicular to the displacement vector, so the dot product will be 0!

 

[suspenc3]

soo...how do i find the work done by gravity

 

[Pyrrhus]

soo...how do i find the work done by gravity

Gravity does work!, but only one component of gravity!, the one that is parallel to the displacement vector.

 

[suspenc3]

but gravity is not perpendicular to the displacement vector..
mgdcos(phi) makes it parallel to the displacement vector doesn't it?

 

[Pyrrhus]

You have the gravity vector pointing down, and if you system of coordinates is directed at the angle the incline has, so the x-axis is parallel to the hypotenuse, and the y-axis is perpendicular to the hypotenuse, then you can decompose gravity in two components, one along the y-axis (mgcos(angle)), and one along the x-axis (mgsin(angle)), the displacement vector goes along the x axis.

 

[suspenc3]

so to find the work done by gravity all i have to do is mgcos(30)

 

[suspenc3]

is everything before this correct by the way?

If so wouldn't Wg equal (-Wf)

 

[Pyrrhus]

so to find the work done by gravity all i have to do is mgcos(30)

No, that's wrong i already explained why.

Yes Wg IS equal -Wf.

 

[suspenc3]

ok..but if i just mark this down i won't get any credit for it

How could I prove that -Wf = Wg using my FBD's?

 

[Pyrrhus]

I already say why
Try reading my #14 and #10 replies.

 

[verty]

The forces are in equilibrium and gravity's working component is the equilibriant of the worker's force. Both of their magnitudes are given by mgsin(arctan(0.9/1.5)).