Physics- Jet Engine Force and Rocket Kinetic Energy Calculation

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A jet engine exerts a force of 300 N on a 50 kg rocket, while air friction applies a force of 200 N, resulting in a net force of 100 N. The kinetic energy gained by the rocket after being thrust horizontally 2.0 m is calculated using the equations of motion and kinetic energy. The correct calculation shows that the final velocity squared (vf^2) equals 8, leading to a kinetic energy (KE) of 200 J. The initial attempt mistakenly used the wrong net force but arrived at the correct kinetic energy value. The discussion confirms that the final answer of 200 J is accurate.
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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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