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Answer :
Final answer:
The total mechanical energy of a 2.00-kg ball dropped into a 10.0-m-deep well just before hitting the bottom is 196 J. This is calculated by applying the principle of conservation of energy to the initial gravitational potential energy, which is completely converted into kinetic energy at the point of impact.
Explanation:
You have asked about the total mechanical energy of a 2.00-kg ball that is dropped into a 10.0-m-deep well just before it hits the bottom, given that it initially has zero kinetic and potential energy. To solve this, we can apply the principle of conservation of energy. Since the ball starts from rest at a height, it has gravitational potential energy given by the formula PE = mgh, where m is the mass, g is the acceleration due to gravity (9.8 m/s²), and h is the height. As the ball falls, this potential energy is converted into kinetic energy, KE = 1/2 mv², where v is the velocity just before it hits the bottom. Therefore just before impact, all the potential energy has been converted to kinetic energy.
Calculating the initial potential energy:
PE = mgh = (2.00 kg)(9.81 m/s²)(10.0 m) = 196.2 J.
Assuming no energy is lost to air resistance or other factors, this amount of energy will be the kinetic energy just before the ball hits the bottom of the well. Therefore, the sum of its kinetic and potential energy just before impact would be equal to the initial potential energy, so:
Option b. 196 J is correct.
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