# Math Help - Work on a inclined plate problem

1. ## 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

2. 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

3. 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

4. maybe they are in another planet or something

5. haha funny....But NOT in this planet .

6. 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 = $-(0.2)8g\cos(30) \cdot 25 = -346.4 \, J$

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 ...

7. from where did that .25 come from?(Nice signature "... work smart, not hard.")

8. Originally Posted by silvercats
from where did that .25 come from?
what 0.25 ? there's a whole number 25 for the displacement ...

9. ahh yeah.I was confused.sorry.I use * for multiplication than . . .Thanks for the help!