Solving Physics Problems: 2.00 kg Box Moving Right

[Rookie8391]

Homework Statement

A 2.00 kg box is moving to the right with a speed of 9.00 m/s, on a horizontal frictionless surface. At t = 0, a horizontal force is applied to the box. The force is directed to the left and has a magnitude F(t) = 6.00 N
A) What distance does the box move from its position at t = 0 before its speed is reduced to zero?
B) If the force continues to be applied what is the speed of the box at t = 3.00s.

Homework Equations

F = ma.
x = (1/2)at² + V_ot + X_o
V_f = at + V_o

The Attempt at a Solution

A) My thinking was that we already know that F = ma. So if F = -6.00N and m = 2.00 kg, then a = -3 (saying the force is being applied in the negative direction.)
Also, we know that V_f = at + V_o, I use this to find how long it takes to stop. 0 = -3t + 9. Here I find t = 3. Lastly, I use x = (1/2)at² + V_ot + X_o to find the distance. x = (1/2)(-3)(3)² + 9(3). Here i get x = 13.5.
B) What I used here was V_f = at + V_o. So - V_f = (-3)(3) + 9, which I get to equal zero.

These were my attemps at solving, but I was wrong, If anyone can help find the problem either with the math or the way of thinking, please let me know, thank you.

 

[gneill]

Homework Statement

A 2.00 kg box is moving to the right with a speed of 9.00 m/s, on a horizontal frictionless surface. At t = 0, a horizontal force is applied to the box. The force is directed to the left and has a magnitude F(t) = 6.00 N
A) What distance does the box move from its position at t = 0 before its speed is reduced to zero?
B) If the force continues to be applied what is the speed of the box at t = 3.00s.

Homework Equations

F = ma.
x = (1/2)at² + V_ot + X_o
V_f = at + V_o

The Attempt at a Solution

A) My thinking was that we already know that F = ma. So if F = -6.00N and m = 2.00 kg, then a = -3 (saying the force is being applied in the negative direction.)
Also, we know that V_f = at + V_o, I use this to find how long it takes to stop. 0 = -3t + 9. Here I find t = 3. Lastly, I use x = (1/2)at² + V_ot + X_o to find the distance. x = (1/2)(-3)(3)² + 9(3). Here i get x = 13.5.
B) What I used here was V_f = at + V_o. So - V_f = (-3)(3) + 9, which I get to equal zero.

These were my attemps at solving, but I was wrong, If anyone can help find the problem either with the math or the way of thinking, please let me know, thank you.

Hi Rookie8391, welcome to Physics Forums.

I don't see any problems with your analysis. Perhaps there's an error in the problem statement itself, or in the answer key.

Is it possible that for (b) they meant to say that the force continues to be applied for 3s after the box achieves zero velocity?