Velocity, Work, and Energy of a Falling Meteor

AI Thread Summary
The discussion revolves around calculating the speed, work, average force, and thermal energy of a meteor falling from a height of 700 km. The initial attempt to find the speed just before impact using the kinematic equation was incorrect due to the assumption of constant acceleration, as gravity varies with altitude. Instead, a conservation of energy approach is recommended, emphasizing that potential energy should not be calculated using the standard mgh formula at such heights. The conversation highlights the importance of recognizing variable gravitational force in high-altitude scenarios. Accurate calculations will lead to the correct answers for subsequent parts of the problem.
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Homework Statement



A meteor has a speed of 94.0 m/s when 700 km above the Earth. It is falling vertically (ignore air resistance) and strikes a bed of sand in which it is brought to rest in 3.03 m.

a) What is its speed just before striking the sand?

b) How much work does the sand do to stop the meteor (mass = 545 kg)?

c) What is the average force exerted by the sand on the meteor?

(d) How much thermal energy is produced?

Homework Equations



KE=0.5*m*v^2

Work=force*distance

vf^2=vi^2+2a(xf-xi)



The Attempt at a Solution



I'm having trouble with the first part, and it looks like the answer for it is required for the other parts.
I tried using vf^2=vi^2+2a(xf-xi) to find the velocity for part a. Plugging in the numbers gives vf^2=(94 m/s)^2+2(9.8 m/s^2)(700000m). This results in vf=3705.244, however this is not correct. What am I doing wrong?
 
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Note that the meteor is falling from significantly higher than ground level. This means that the force of gravity will not be constant when it is falling. The equation you are attempting to use, v_f^2 = v_i^2 + 2A \Delta x, is meant to be applied to problems where the acceleration will be constant. I would think this is your mistake. I would suggest trying part a as a conservation of energy problem. Keep in mind that the potential energy of gravity will not be mgh for the same reasons I pointed out above.
 
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