# Work Done?

1. Dec 11, 2003

### dzem68

Work Done???

If someone has the time and patience could you help me in setting this problem up?

300kg object moves 4.5 meters down a 25 degree incline at a constant speed. It is kept from accelerating by a force pushing back on it. The effective coefficient of friction is .39

I need to:

a) calculate the net work done on the object
b) calculate the work done by the force pushing back on it
c) calculate the work done by gravity on the object.

I do not know where to begin. could someone point me in the right direction.

thanks,

dz

2. Dec 11, 2003

### ShawnD

Re: Work Done???

gravity force:
Fg = 300 * 9.81
Fg = 2943N

normal force:
N = Fgcos(theta)
N = 2943cos(25)
N = 2667N

friction force:
f = uN
f = 0.39 * 2667
f = 1040N

Since the thing is not accelerating, net forces are 0. I drew a FBD of the box with the X axis along the 25 degree slant and the Y is the same direction as the normal force.
sum of x = 0 = 1040 + F - 2943sin(25)
F = 203.8N (the force preventing it from accelerating)

I don't exactly know what you mean by "net work". If I had to guess, I would say the net work is 0 since the mass is not accelerating.

work done by force:
W = Fd
W = 203.8 * 4.5
W = 917.1J

work done by gravity:
W = Fd
W = 2943sin(25) * 4.5
W = 5597J

Last edited: Dec 11, 2003
3. Dec 11, 2003

### dzem68

Thanks ShawnD! I read through your respons and I think I actually understand the concepts!!