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Answer :
Final answer:
The stone, after the flexible pole does work on it, leaves with a speed of about 26.02 m/s. This is calculated using the work-energy theorem and the given values for the mass of the stone, its initial speed, and the work done by the pole.
Explanation:
The subject of the question is physics, focusing particularly on the principles of kinetic energy and work. We need to apply the work-energy theorem, which states that the work done on an object is equal to the change in its kinetic energy.
Initial kinetic energy of the stone can be calculated using the formula for kinetic energy KE = 0.5 * m *v^2, where m is mass and v is velocity. So initial kinetic energy is 0.5 * 0.176kg * (37.9 m/s)^2 = 134.15244 J.
The work done by the pole on the stone is given as -58.3 J (the work is negative because it is done against the direction of motion of the stone). According to the work-energy theorem, this work is equal to the change in kinetic energy. Therefore, the final kinetic energy of the stone is 134.15244 J - 58.3 J = 75.85244 J.
The final speed of the stone can be calculated by rearranging the kinetic energy formula to solve for v: v = sqrt((2*KE)/m). Substituting the given values, we get v = sqrt((2*75.85244 J)/0.176kg) = 26.02 m/s.
Therefore, after the pole hits the stone, it leaves with a speed of roughly 26.02 m/s.
Learn more about Work-Energy Theorem here:
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