29 Sep '08 00:07>1 edit
I need help with the logic , I don't need the answer...
So,....How much work is done by the horizontal force, if force of the pull = 150N on the 18kg block when the force pushes the block 5.0meters up the 32degree incline. The coefficient of friction is .10
And the force is parallel to the adjacient side of the inclination.
My actual question is: Why is the work done in that direction simply the force of the push, times the distance, times the cos of the angle between them, when i can't seem to escape the notion that it should be the net force of the push, times the distance, times the angle between them?
by net force I mean Fp*cos(theta))-mk*[(m*g*cos(theta) + Fp*sin(theta)]-(m*g)*sin(theta)
Fp=Force push
mk= coefficient of friction
m=mass
g= acceleration due to gravity
So,....How much work is done by the horizontal force, if force of the pull = 150N on the 18kg block when the force pushes the block 5.0meters up the 32degree incline. The coefficient of friction is .10
And the force is parallel to the adjacient side of the inclination.
My actual question is: Why is the work done in that direction simply the force of the push, times the distance, times the cos of the angle between them, when i can't seem to escape the notion that it should be the net force of the push, times the distance, times the angle between them?
by net force I mean Fp*cos(theta))-mk*[(m*g*cos(theta) + Fp*sin(theta)]-(m*g)*sin(theta)
Fp=Force push
mk= coefficient of friction
m=mass
g= acceleration due to gravity