# Homework Help: Honors physics intermediate force question

1. Jun 21, 2011

### Kalix

1. The problem statement, all variables and given/known data
Question: An object traveling at a constant speed of 22m/s begins to experience a force of 15N in the opposite direction of its motion. (Hint given: Start by finding objects acceleration)

a. How long will it take for the 8kg object to come to a complete stop once it experiences the force?
b. If the object was traveling in a straight line, how far did the object travel as it slowed down?

2. Relevant equations
I am not exactly sure what the relevant equations are for this problem but here are the equations we have used so far in this unit.

F=ma
W=mg
Fs=μsxFn (Static friction=mu times normal force)
Fk=μkxFn (Kinetic friction=mu times normal force)

I am guessing that I have to use one of those equations.
3. The attempt at a solution
Here is my small attempt.
F=ma
15N=(8g)(a)
a=1.875m/s^2

This is where I get stuck. First off I don't even know if the value I got for acceleration is correct and secondly where do I go from here. Do I go back to my kinematics equations to find time in part "a" and X in part "b"?

2. Jun 21, 2011

### Staff: Mentor

Your acceleration is correct. And yes, time to dust off the kinematics equations to answer the questions.

3. Jun 21, 2011

### Kalix

Would it be easier to find time (t) first or X first? And I know that the final velocity will be 0m/s but what is the initial velocity. I don't think it's 22m/s but I guess it could be...

4. Jun 21, 2011

### Staff: Mentor

Sure it is. That's given.

5. Jun 21, 2011

### Kalix

I ended up getting a negative value for X...is that possible?

6. Jun 21, 2011

### Staff: Mentor

No. Show what you did.

Did you find the time first?

7. Jun 21, 2011

### Kalix

No I found X first. This is what I did.
Vfx^2=Vix^2+2aX
0=(22^2)=2(1.875)(X)
-484=3.75(x)
x=-129.07

X=Vxt
-129.07=22(t)
t=-5.87sec

What did I do wrong?

8. Jun 21, 2011

### Staff: Mentor

You have the wrong sign for the acceleration. Remember it acts opposite to the velocity, so if the velocity is positive the acceleration must be negative.