# Homework Help: Newtons first and second laws

1. Sep 19, 2008

### ghostrider989

1. A 3600 jet touches down at 250 on the deck of an aircraft carrier and immediately deploys a parachute to slow itself down. If the plane comes to a stop in 180 , what is the average force of air on the parachute? Assume the parachute provides essentially all the stopping force.

2. Fnet= mass*acceleration, a=v/t, v=p/t

3. M=3600kg
v=250km/h
p=180

v=p/t 180m/25000m/h convert the kilometers to meters t=7.2e^-4*3600
t=2.592

250000m(1/3600sec)= 69.4 m/s

fnet=ma

m=3600

fnet=3600kg(69.4(m/s)/(2.592 s)= 96450N

i tried doing it but i keep on getting it wrong.

2. Sep 20, 2008

### Kurdt

Staff Emeritus
3. Sep 20, 2008

### ghostrider989

]1. A 3600kg jet touches down at 250km/h on the deck of an aircraft carrier and immediately deploys a parachute to slow itself down. If the plane comes to a stop in 180m , what is the average force of air on the parachute? Assume the parachute provides essentially all the stopping force.

I was thinking about using v2=v02+2a deltat

4. Sep 20, 2008

### ghostrider989

am i on the right track if i do that

5. Sep 20, 2008

### Mattowander

If you're asking whether or not using the equation v squared = v0 squared + 2as would help you, then yes you would be on the right track.

Another way to solve this would be by using the impulse momentum theorem Ft = change in momentum.

Hope this helps!

6. Sep 20, 2008

### Kurdt

Staff Emeritus
Yes you can use that equation to find acceleration.

7. Sep 20, 2008

### cepheid

Staff Emeritus
Wouldn't it make more sense to use the work-energy theorem? I mean, you're given the distance over which the force acts, not the time interval.

8. Sep 20, 2008

### Mattowander

Now that I think about it yes :)

Just goes to show that there are many different ways to solve a given problem.