# Work Question

1. Nov 1, 2005

### 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

Last edited: Nov 1, 2005
2. Nov 1, 2005

### 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

3. Nov 1, 2005

4. Nov 1, 2005

### 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.

5. Nov 1, 2005

### 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

6. Nov 1, 2005

### Pyrrhus

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

7. Nov 1, 2005

### 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 cant be correct because Wg +Wf = 0?

8. Nov 1, 2005

### Pyrrhus

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

9. Nov 1, 2005

### suspenc3

1.5 = hypotenuse
0.9 = height

10. Nov 1, 2005

### 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!

11. Nov 1, 2005

### suspenc3

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

12. Nov 1, 2005

### Pyrrhus

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

13. Nov 1, 2005

### suspenc3

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

14. Nov 1, 2005

### 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.

15. Nov 1, 2005

### suspenc3

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

16. Nov 1, 2005

### suspenc3

is everything before this correct by the way?

If so wouldnt Wg equal (-Wf)

17. Nov 1, 2005

### Pyrrhus

No, that's wrong i already explained why.

Yes Wg IS equal -Wf.

18. Nov 1, 2005

### suspenc3

ok..but if i just mark this down i wont get any credit for it

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

19. Nov 1, 2005