# 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.

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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