Work Done???

[dzem68]

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

[ShawnD]

Originally posted by dzem68
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.



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

[dzem68]

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