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Question: A .08Kg ball is dropped from a height of 3m. The ball bounces back up to a height of 1.2m. How much mechanical energy was transferred from the ball to the environment?
To solve this problem, we have to first calculate the initial mechanical energy before the ball was dropped and the final mechanical energy after the ball has bounced back. The mechanical energy here is just the gravitational potential energy because the ball is initially at rest and finally also at rest momentarily. The potential energy (PE) can be given by the equation PE = m*g*h, where m is mass, g is acceleration due to gravity (usually assumed to be 9.8m/s^2 unless otherwise stated), and h is height. 1. Initial mechanical energy (before the drop): PE initial = m * g * h = 0.08 kg * 9.8 m/s^2 * 3 m = 2.352 J (joules) 2. Final mechanical energy (after the bounce): PE final = m * g * h = 0.08 kg * 9.8 m/s^2 * 1.2 m = 0.9408 J The mechanical energy transferred to the environment is the difference between the initial and final mechanical energy. Energy transferred = PE initial - PE final = 2.352 J - 0.9408 J = 1.4112 J So, approximately 1.41 Joules of mechanical energy were transferred from the ball to the environment in the process of bouncing.
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