# Newton's Laws, Force and Kinematics

1. Jul 27, 2009

### skydiver

1. The problem statement, all variables and given/known data
A particle is traveling in a straight line at a constant speed of 24.5 m/s. Suddenly, a constant force of 14.3 N acts on it, bringing it to a stop in a distance of 52.9 m.

a. Determine the time it takes for the particle to come to a stop.

b. What is its mass?

2. Relevant equations

1. vx2 = v0x2 +2a$$\Delta$$x

2. $$\Delta$$x = v0t + 0.5at2

3. F=ma

3. The attempt at a solution

I know how to solve part a. I use equation 1 to find a which is 5.673 m/s^2
I use that value of a in equation 2 to find t which is 4.318 s
This is all correct.

I am having trouble with part b - to find mass.
The only equation I have to find mass is f=ma
F is 14.3N given, a as calculated is 4.318 which gives m as 3.311 kgs which is incorrect.
The correct answer is 2.52 kg

So I thought that for F=ma, F must be the resultant force of some sort.
So I subtracted 24.5-14.3 (which I am sure is not a correct step to do) gives me 10.2

And that divided by a (4.318) gives me m as 2.362 which is close but not correct.

Any other thoughts? Thanks so much

EDIT: Aarggghh.. stupid me
I am diving by t (4.318) instead of the a I found (5.673).
Can't believe I spent so much time on this stupid stupid mistake.

2. Jul 27, 2009

### tiny-tim

Welcome to PF!

Hi skydiver! Welcome to PF!

I'm glad you've sorted it out!

Just for the record, there's another way of finding the mass …

just calculate the work done and then use the work-energy theorem … work done = change in energy.

3. Jul 27, 2009

### RoyalCat

Re: Welcome to PF!

The really beautiful thing is, that the first equation he posted:
$$v_f^2=v_0^2+2ad$$ is completely equivalent.
Multiplying by a factor of $$\tfrac{1}{2} m$$ provides:

$$\tfrac{1}{2} mv_f^2=\tfrac{1}{2} mv_0^2+Fd$$

(The dot operator is included in the way the equation is defined, since the $$a$$ term refers to the acceleration in the direction of the distance traveled.)