Newtons first and second laws

[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.

 

[Kurdt]

It would help enormously if your numbers had units in the question, otherwise nothing makes sense.

You need to work out the acceleration using the kinematic equations and then use F= ma.

https://www.physicsforums.com/showpost.php?p=905663&postcount=2

 

[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

 

[ghostrider989]

am i on the right track if i do that

 

[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!

 

[Kurdt]

Yes you can use that equation to find acceleration.

 

[cepheid]

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!

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.

 

[Mattowander]

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.

Now that I think about it yes :)

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