# Help Again Please - Box sliding on a floor

• jbow615
In summary, the problem involves a box with a mass of 12 kg and a friction coefficient of 0.23, traveling a distance of 4 meters to the east. The four forces acting on the box are 10 N to the east, 10 cos(50) N at an angle of 50 degrees above the horizontal towards the east, 10 N to the south, and 10 N to the west. In order to find the change in kinetic energy, we must calculate the friction force and the applied force, and then subtract the friction force from the applied force. The work done by friction is 108.942 joules, while the work done by the applied force is 25.711 joules. Therefore

#### 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.

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

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?