# Frictional Forces acting upon a moving object

1. Feb 16, 2010

### HaXxZorZ_X

1.A 22 kg box is being pushed across the floor by a constant force ‹ 100, 0, 0 › N. The coefficient of kinetic friction for the table and box is 0.15. At t = 8.0 s the box is at location ‹ 13, 3, −2 › m, traveling with velocity ‹ 4, 0, 0 › m/s. What is its position and velocity at t = 9.3 s in terms of an < x , y , z > vector?

2. Δpx = -µk*FNΔt

Δvx = -µk*g*Δt

3. I used the formula Δvx = -µk*g*Δt and tried to solve for Vf.
Vfx - 4 = -(.15)*(9.8)*(9.3 - 8)
Vfx = -(.15)*(9.8)*(9.3 - 8) + 4
Vfx = 2.089

According to my textbook that answer should be correct, but its not. Does anyone have any tips to help me figure out what im doing wrong, or what i should do different?

I also understand that the y and z components of the velocity are not going to change because the force is only in the x direction. So the velocity vector is going to be something like < x , 0 , 0 > i just cant figure out how to find the x value of the velocity.