# What is The New velocity of the box?

1. Sep 29, 2010

### ohheytai

1. The problem statement, all variables and given/known data
A 24 kg box is being pushed across the floor by a constant force ‹ 107, 0, 0 › N. The coefficient of kinetic friction for the table and box is 0.18. At t = 8.0 s the box is at location ‹ 11, 2, −4 › m, traveling with velocity ‹ 5, 0, 0 › m/s. What is its position and velocity at t = 9.7 s?

2. Relevant equations

3. The attempt at a solution
i got <12.11,0,0> and <7.58,0,0> but they were both wrong

2. Sep 29, 2010

### rock.freak667

Since the force is in the x-direction, you need to get the frictional force which would be Frictional force = <μmg,0,0>, then find the resultant force,F. (and F=ma )

3. Sep 29, 2010

### ohheytai

yea but i still dont get how to find v final at 9.7s

4. Sep 29, 2010

### rock.freak667

Using my suggestion, once you get the resultant force vector and apply F=ma, you can get the vector a.

Then you can use v(t) = ∫a(t) dt

5. Sep 29, 2010

### ohheytai

can you show me how to get the answer?

6. Sep 29, 2010

### rock.freak667

I can't pose the solution for you, but I will start you off.

Resultant force, ma(t) = Constant force - frictional force