# Work on a inclined plate problem

• May 25th 2011, 02:49 AM
silvercats
Work on a inclined plate problem
8kg object is dragged 25m distance on a 30 degrees
inclined plate.if the friction coefficient is 0.2 between the plate and the object,
what is the work done agaisnt the friciton and the gravity?

I think i found the work against the frictions which is wrong and NOT the answer at the behind of book
80N*Sin 30 =40N <<is the force on the inclined plate of the 80N object
so the Friction is 0.2*40 =8N

so work against the friction is 25*8 =200N which is wrong

work agianst the Gravity is:
sin 30*25 =12.5 m which is vertical distance traveled.
so Work=F*D =80N*12.5m = 1000
• May 25th 2011, 09:30 AM
bugatti79
Quote:

Originally Posted by silvercats
8kg object is dragged 25m distance on a 30 degrees
inclined plate.if the friction coefficient is 0.2 between the plate and the object,
what is the work done agaisnt the friciton and the gravity?

I think i found the work against the frictions which is wrong and NOT the answer at the behind of book
80N*Sin 30 =40N <<is the force on the inclined plate of the 80N object
so the Friction is 0.2*40 =8N

so work against the friction is 25*8 =200N which is wrong

work agianst the Gravity is:
sin 30*25 =12.5 m which is vertical distance traveled.
so Work=F*D =80N*12.5m = 1000

The force required to drag the object up the plane is a combination of the friction force and the force due to gravity.
The friction force is F=u*n where n is the normal force acting perpendicular to the inclined plane and is
F1=0.2*8*9.81*cos 30 = 13.593N

The force due to gravity is F2=w*sin 30 = 39.24N

Total force is F1+F2=52.833N

Work done is W=F*d=52.833N*25m=1321Nm
• May 26th 2011, 07:15 AM
silvercats
Quote:

Originally Posted by bugatti79
The force required to drag the object up the plane is a combination of the friction force and the force due to gravity.
The friction force is F=u*n where n is the normal force acting perpendicular to the inclined plane and is
F1=0.2*8*9.81*cos 30 = 13.593N

The force due to gravity is F2=w*sin 30 = 39.24N

Total force is F1+F2=52.833N

Work done is W=F*d=52.833N*25m=1321Nm

but book says work against the gravity is 346.4j
and work against the gravity is 1000j :-O
• May 26th 2011, 08:44 PM
igcsestudent
maybe they are in another planet or something :p
• May 26th 2011, 09:01 PM
silvercats
haha funny....But NOT in this planet .
• May 27th 2011, 09:57 AM
skeeter
Quote:

8kg object is dragged 25m distance on a 30 degrees
inclined plate.if the friction coefficient is 0.2 between the plate and the object,
what is the work done agaisnt the friciton and the gravity?
using 10 for g ...

work done by gravity = $-(mgh) = -8g(12.5) = -1000 \, J$

work done by friction = $-(0.2)8g\cos(30) \cdot 25 = -346.4 \, J$

Quote:

but book says work against the gravity is 346.4j
and work against the gravity is 1000j :-O
I think you meant work against friction first ...
• May 27th 2011, 05:12 PM
silvercats
from where did that .25 come from?(Nice signature "... work smart, not hard.")
• May 28th 2011, 01:06 PM
skeeter
Quote:

Originally Posted by silvercats
from where did that .25 come from?

what 0.25 ? there's a whole number 25 for the displacement ...
• May 28th 2011, 06:22 PM
silvercats
ahh yeah.I was confused.sorry.I use * for multiplication than . .:D .Thanks for the help!