# a question involving physics

joe shmo
Posers and Puzzles 29 Sep '08 00:07
1. joe shmo
Strange Egg
29 Sep '08 00:071 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
2. 01 Oct '08 03:34
The net force will perform work which creates kinetic energy for the moving mass. But there is also work to be performed to overcome the friction (the incline and the mass heat up somewhat) and finally work is performed to increase the potential energy of the mass: It is 5 m * sin(32 degrees) higher once it travelled 5 meters over the incline. That is - if I understood the description correctly.
3. 01 Oct '08 03:56
In your formula, the net force works over a distance of d = 5 meter and accelerates the mass to a kinetic energy of
d * (Fp*cos(theta))-mk*[(m*g*cos(theta) + Fp*sin(theta)]-(m*g)*sin(theta))

The amount of friction heat induced by the partial gravity (reduced by the incline) and also by the force Fp partially pulling the mass against the incline is
d * mk * [(m*g*cos(theta) + Fp*sin(theta)]

And the potential energy gained is d * (m*g) * sin(theta) - as the incline carries most of the weight.

Sum them and you get what you originally said the work was: d * Fp * cos(theta), i.e. all performed by the moving force along the incline.