Quick question on force

  • #1
Here is the problem:

Calculate the force exerted on a rocket, given that the propelling gases are expelled at a rate of 1500kg/s with a speed of 4*10^4 m/s (at the moment of take off)

I know that force is equal to the mass times the change in velocity over the change in time. Can I use a one second interval for the time and just make up a time like that?
 

Answers and Replies

  • #2
574
2
Yes, you are on the right track. The gas exerts the same force on the rocket as the rocket exerts on the gas. Find the acceleration of the gas by setting time equal to one second, and displacement equal to [itex]4\times10^{4}[/itex] meters. Then use [itex]F_{net}=ma[/itex].
 
  • #3
Solving for acceleration won't the acceleration be the same as the velocity in this equation?? I used the formula of Vf=Vo+at
 
  • #4
Andrew Mason
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Flinthill84 said:
Here is the problem:

Calculate the force exerted on a rocket, given that the propelling gases are expelled at a rate of 1500kg/s with a speed of 4*10^4 m/s (at the moment of take off)

I know that force is equal to the mass times the change in velocity over the change in time. Can I use a one second interval for the time and just make up a time like that?
Use:

[tex]\vec F = \frac{\delta \vec p}{\delta t}[/tex]

In this case:

[tex]\frac{\delta p}{\delta t} = v \frac{\delta m}{\delta t}[/tex]

Where [itex]\frac{\delta m}{\delta t} = 1500 kg/sec[/tex]

AM
 
  • #5
Thankyou very much!!!
 
  • #6
574
2
Sirus said:
Yes, you are on the right track. The gas exerts the same force on the rocket as the rocket exerts on the gas. Find the acceleration of the gas by setting time equal to one second, and displacement equal to [itex]4\times10^{4}[/itex] meters. Then use [itex]F_{net}=ma[/itex].
Oh boy...forget this post. Not sure what I was thinking here... :eek:
 

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