(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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.

2. Relevant equations

F = ma.

x = (1/2)at^{2}+ V_{o}t + X_{o}

V_{f}= at + V_{o}

3. 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)at^{2}+ V_{o}t + 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.

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# Force on a Box

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