What is the net force on a weather rocket during its flight?

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SUMMARY

The net force on a weather rocket during its flight can be calculated using Newton's second law, specifically the modified form that accounts for changing mass. Given the rocket's mass of 48 kg, an acceleration of 18 m/s², and a mass loss rate of 0.5 kg/s, the net force can be determined using the equation F = dp/dt, where p is momentum. The correct application of the product rule for differentiation yields the necessary insights to solve the problem accurately.

PREREQUISITES
  • Understanding of Newton's second law of motion
  • Familiarity with the concept of momentum (p = mv)
  • Knowledge of calculus, specifically differentiation
  • Basic principles of rocket propulsion and mass flow rate
NEXT STEPS
  • Study the product rule in calculus for differentiating products of functions
  • Learn about the implications of changing mass in rocket dynamics
  • Explore advanced applications of Newton's laws in variable mass systems
  • Investigate real-world examples of weather rockets and their operational physics
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Students in physics, aerospace engineers, and anyone interested in the dynamics of rocket flight and the application of Newton's laws in variable mass scenarios.

bcjochim07
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Homework Statement


A small rocket to gather weather data is launched straight up. Several seconds into the flight, its velocity is 120 m/s and it is accelerating at 18 m/s^2. At this instant, the rocket's mass is 48 kg and it is losing mass at the rate of .5 kg/s as it burns fuel. What is the net force on the rocket? Hint: Newton's second law was presented in a new form in this chapter


Homework Equations


p=mv F=dmv/dt F=dp/dt


The Attempt at a Solution



I tried to come up with a formula for momentum with respect to time and then differentiate it to find force, but it's not making sense to me.

I tried to say that at this instant dm/dt= -.5 and dv/dt = 18 and then I tried to multiply those but that clearly is not correct reasoning. I'm pretty sure I'm not looking at this problem right. Any hints?
 
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Welcome to PF!

bcjochim07 said:
I tried to say that at this instant dm/dt= -.5 and dv/dt = 18 and then I tried to multiply those but that clearly is not correct reasoning. I'm pretty sure I'm not looking at this problem right. Any hints?

Hi bcjochim07 ! Welcome to PF! :smile:

A hint? :rolleyes: … yes! …

Product rule: d(mv)/dt = mdv/dt + vdm/dt. :smile:
 
Oh yes, I see. Thank you very much!
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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