Is the Work Done by Gravity Calculated Incorrectly?

AI Thread Summary
The discussion revolves around a physics problem involving a block sliding down an incline with an applied force. The user calculated the work done by the applied force as 401J and the work done by gravity as -330.8J, questioning the relationship between these values. It was clarified that the work done by gravity should only consider the component of gravity parallel to the displacement, leading to confusion about the calculations. The forces acting on the block are in equilibrium, indicating that the work done by gravity is equal to the negative of the work done by the applied force. The conversation emphasizes the importance of understanding the components of forces and their impact on work calculations in physics.
suspenc3
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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
 
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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
 
Please post the problem statement!
 
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.
 
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
 
Ok so show me your work, and/or where did you get stuck?
 
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?
 
i suppose the 1.5 and 0.9 are sides of the triangle right?
 
1.5 = hypotenuse
0.9 = height
 
  • #10
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!
 
  • #11
soo...how do i find the work done by gravity
 
  • #12
suspenc3 said:
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.
 
  • #13
but gravity is not perpendicular to the displacement vector..
mgdcos(phi) makes it parallel to the displacement vector doesn't it?
 
  • #14
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.
 
  • #15
so to find the work done by gravity all i have to do is mgcos(30)
 
  • #16
is everything before this correct by the way?

If so wouldn't Wg equal (-Wf)
 
  • #17
suspenc3 said:
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.
 
  • #18
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?
 
  • #19
I already say why :rolleyes:
Try reading my #14 and #10 replies.
 
  • #20
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)).
 

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