Energy conservation answer check

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SUMMARY

The discussion centers on calculating the displacement along a ramp for a 1500kg rocket launched with an initial velocity of 50m/s and a constant thrust of 2000 N, while overcoming a friction force of 500 N. The ramp is inclined at 53 degrees. The user applied the energy conservation equation, E0 + Win = Ef + Elost, and determined the height using U = mgh and the kinetic energy formula KE = 1/2 m v². The calculated displacement along the ramp is 141.6 meters, confirming the solution's accuracy.

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  • Understanding of energy conservation principles in physics
  • Familiarity with basic kinematic equations
  • Knowledge of trigonometric functions, specifically sine
  • Ability to apply Newton's laws of motion
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  • Review energy conservation in mechanical systems
  • Explore the application of trigonometry in physics problems
  • Study the effects of friction on motion
  • Learn about the dynamics of rocket propulsion
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Homework Statement



A 1500kg rocket is to be launched with an initial upward velocity of 50m/s. In order to assist its engines, engineers will start it from rest on a ramp that rises 53 degrees from the horizontal. At the bottom, the ramp turns upward and launches the rocket vertically. The engines provide a constant forward thrust of 2000 N, and friction with the ramp is a constant 500 N. How far from the base of the ramp should the rocket start, as measured along the surface of the ramp?

Homework Equations



E0 +Win = Ef + Elost
U = mgh
KE = 1/2 m v2

The Attempt at a Solution



I need to find the displacement along the ramp. I made the change in x the surface of the ramp, and I made h equal x*sin 53.

Solving for x, I got 141.6m.
 
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