# Laws of motion HELP

1. Jun 8, 2005

### mamelancon84

Been struggling wih this problem all afternoon.

On an aircraft carrier, a jet can be catapulted fom 0 to 155 mi/hr in 2.00 sec. If the average force exerted by the catapult is 5.10 * 10^6 Newtons, what is he mass of the jet?

We converted the velocity into meters/second and got 69.276 meters/second

initial velocity = 0
initial X = 0
final X = (initial X) + (velocity of X)(time)
= 0 + 69.276(2.00)
= 138.6 meters

We don't know where to go from here...Summer classes. Test every Friday!!! NEED HELP ASAP!
Thanks

2. Jun 8, 2005

### whozum

Write down everything the problem gives you.
F = 5.1 * 10^6 N
m = ?
$\Delta$V = 69.276m/s
$\Delta$t = 2s

What do you think comes next? You're trying to find M.

3. Jun 8, 2005

### Pyrrhus

Ok, you know a force made by the catapult produces an acceleration, so the jet will get from 0 to 155 in 2 seconds, you can find the acceleration (throught its definition) and then use Newton's 2nd Law.

4. Jun 8, 2005

### mamelancon84

That's too simple. If we were to use F=ma, acceleration would most likely be -9.8 m/s^2. then that would be 5.10 *10^6N/ -9.8m/s^2

But what is the time used for???
We were given this equation: X(final) = X(initial) + velocity of x * time + .5(acceleration)(time^2)

Where we are confused is we were never given an angle that it was catapulted, or a distance to go by, so we are stuck on this next step?!?

5. Jun 8, 2005

### mamelancon84

The formula was introduced earlier in the chapter, and it is something we have been using in other problems. I assume that is what we are supposed to be using.

6. Jun 8, 2005

### Pyrrhus

You believe the airplane doesn't run a horizontal trajectory before taking off?, This problem, indeed, has several simplifications, i imagine the purpose was what Whozum and i suggested.

7. Jun 8, 2005

### whozum

You need to use the impuls equation which I hinted at already. Acceleration would not be 9.8m/s^, it would be 69.276/2 = 34.638m/s^2.

$$F\Delta t = m\Delta v$$

edit: clarificatoin

Newton's 2nd Law

$$F = ma$$

Average acceleration can be expressed as

$$a = \frac{\Delta v}{\Delta t}$$ which follows directly from it's definition. Plug that into the above to get what I just gave you.

8. Jun 8, 2005

### mamelancon84

Thank you so much! Really appreciate it...don't you hate when its staring you in the face :(