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The question posted at the link provided states that a rocket moving upwards at 210 m/s drops a segment with mass 3 kg at a height 7100 m. The kinetic energy of the segment is to be determined when it reaches the ground.
Here, the acceleration due to gravity, equal to 10 m/s^2, acts in a vertically downwards direction and the rocket is moving upwards at 210 m/s. The direction in which the rocket is moving is opposite to the direction of acceleration due to gravity.
This has been taken into account in the kinetic energy determined. To clarify, let us derive the final kinetic energy using a different method. The total energy of the segment at the height of 7100 m is equal to its total energy at ground level.
The total energy of the segment is the sum of its kinetic energy and its gravitational potential energy. At a height of 7100 m, the gravitational potential energy is m*g*h = 3*10*7100 = 213000 J and the kinetic energy is (1/2)*3*210^2 = 66150 J. The total energy is 279150 J
At ground level, the gravitational potential energy is 0, and the entire energy in the segment is in the form of kinetic energy. This is equal to 279150 J which is the same as the value determined in the earlier question.
Posted by justaguide on August 21, 2013 at 6:04 PM (Answer #1)
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