# Thread: Work Force in Joules

1. ## Work Force in Joules

might need help just setting up a diagram, with the angles, i believe i n need that.

Then i can use the equation W = (F*cos theta) * s

where s is the distance.

thanks for any help with this one.

heres the problem:
A 44 kg box is being pushed a distance of 7.0 m across the floor by a force whose magnitude is 157 N.

And the force is parallel to the displacement of the box.
The coefficient of kinetic friction is 0.25.

I must use the proper + and - signs for the work by each force of:

applied force: in Joules
frictional force: in Joules
normal force: in Joules
gravity: in Joules

also, since i have different Joules for each force, the equation should change, so what variable would be the changing variable in the equation?

thanks alot for any extra help for this one.

2. Originally Posted by rcmango
might need help just setting up a diagram, with the angles, i believe i n need that.

Then i can use the equation W = (F*cos theta) * s

where s is the distance.

thanks for any help with this one.

heres the problem:
A 44 kg box is being pushed a distance of 7.0 m across the floor by a force whose magnitude is 157 N.

And the force is parallel to the displacement of the box.
The coefficient of kinetic friction is 0.25.

I must use the proper + and - signs for the work by each force of:

applied force: in Joules
frictional force: in Joules
normal force: in Joules
gravity: in Joules

also, since i have different Joules for each force, the equation should change, so what variable would be the changing variable in the equation?

thanks alot for any extra help for this one.
Work done by a constant force over a given displacement is given by:
$W = \vec{F} \cdot \vec{s} = Fs~cos(\theta)$
where F and s are the magnitudes of the associated vectors and $\theta$ is the angle between the force and the displacement.

So for example, consider the applied force: This force is parallel to the (level) floor, so logic tells us that the applied force is in the direction of the displacement. Thus the angle between the applied force and the displacement is 0 degrees. Thus
$W_{applied} = \vec{F_{app}} \cdot \vec{s} = F_{app} s ~ cos(0^o) = F_{app} s$

You do the rest. (Yes it is possible for a force to do no work, or even negative work.)

-Dan