Physics- Jet Engine Force and Rocket Kinetic Energy Calculation

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SUMMARY

The discussion focuses on calculating the kinetic energy gained by a 50 kg rocket subjected to a net force of 100 N (300 N thrust minus 200 N air friction) over a distance of 2.0 m. The relevant equations include Newton's second law (f=ma), the kinematic equation (vf^2=vi^2+2ad), and the kinetic energy formula (KE=1/2mv^2). The correct kinetic energy calculation confirms that the rocket gains 200 J of kinetic energy, validating the final answer provided by the participant.

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  • Understanding of Newton's second law (f=ma)
  • Familiarity with kinematic equations (vf^2=vi^2+2ad)
  • Knowledge of kinetic energy calculations (KE=1/2mv^2)
  • Basic principles of forces and motion in physics
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Physics--please check answer

Homework Statement



A jet engine applies force of 300 N horizontally to the right against a 50 kg rocket. The force of air friction 200 N. If the rocket is thrust horizontally 2.0 m ,what is the kinetic energy gained by the rocket?

Homework Equations



I think these equations would be used:
f=ma

vf^2=vi^2+2ad

KE=1/2mv^2

The Attempt at a Solution



First I re-arranged the force formula and got a=f/m. Then I substituted (300-100)for f and 50 for m. I simplified it to 2.
Then I used vf^2=vi^2+2ad and substituted 0 for vi, 2 for a and 2.0 for d. Then I got 2.83 for vf.
Lastly, I used KE=1/2mv^2 and substituted 50 kg for m and 2.83 for v. My final answer is 200 J. I am not too sure if it is correct. Thanks in advanced!
 
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Looks right.
You meant (300-200) above, not (300-100) but you did it right anyway.
There was no need to solve for vf. KE = 1/2 m vf^2 and vf^2 = 8. 8*50/2 = 200.
 
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