Help Again Please - Box sliding on a floor

[jbow615]

Help Again Please -- Box sliding on a floor

Homework Statement

If the friction coefficient between the box and the floor is .23, and the box has a mass of 12 kg, and the magnitude of all four forces is 10 N, what is the change in kinetic energy of the box after it has traveled a distance of 4 meters to the east?

Homework Equations

Friction force =(mu)(normal force)
w total= Kinetic final - Kinetic Initial
W total= force * distance
kinetic energy is 1/2mv^2

The Attempt at a Solution

The attempt was calculating the friction force, calculating the applied force, and then subtracting the friction force from the applied force.

 

[phinds]

Show your work.

 

[jbow615]

Friction force=(mu)(n). N=mg=(12)(9.8)=(117.6)(.23)=27.048
Work done by friction=(27.048)(4)=108.942

And I forgot this part of the question:
4 forces act on a crate on a friction less surface. Force 1 acts due east at 10 N, force 2 at at angle 50 degrees above the horizontal towards the east, force 3 due south, and force 4 due west. What is the work done by each force in increasing order?

So work done by applied force = 10cos(50)*4=25.711

So change in Kinetic Energy= Work of applied force - work of friction force which equals 25.711-108.942=-82.48

 

[haruspex]

You seem to have conflated two questions. In the OP there is friction, in your second post there is none. The OP asked for change in KE of the box, while the second post asks for the work done by each force.
I don't understand your attempt at an answer. Where does the 4 come from? You would need to be multiplying a force by a distance; what distance?