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A 2 kg ball is dropped from a height of 5 m. It hits the ground with a speed of 8 m/s. How much work was done by air resistance as it fell?

a. 15 J
b. 0 J
c. −15 J
d.34 J
e. −34 J

User Cetin Sert
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1 Answer

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Final answer:

The work done by air resistance as the ball fell is calculated by subtracting its final kinetic energy from its initial potential energy, resulting in -34 J. This means that air resistance did negative work, removing energy from the system.

Step-by-step explanation:

To determine how much work was done by air resistance, we can use the work-energy principle. The work done by non-conservative forces, such as air resistance, can be calculated by comparing the mechanical energy at the beginning and the end of the ball's fall.

First, we find the gravitational potential energy at the start (Ui) and the kinetic energy at the end (Kf). The potential energy is Ui = mgh, where m is the mass, g is acceleration due to gravity (9.8 m/s2), and h is the height. The kinetic energy is Kf = 1/2 mv2, where v is the final velocity.

The ball's initial potential energy is Ui = (2 kg)(9.8 m/s2)(5 m) = 98 J. Its final kinetic energy is Kf = 1/2 (2 kg)(8 m/s)2 = 64 J.

If there were no air resistance, the initial potential energy would equal the final kinetic energy. The work done by air resistance can be found by subtracting the final kinetic energy from the initial potential energy: Workair = Kf - Ui = 64 J - 98 J = -34 J.

The negative sign indicates that the work done by air resistance is in the opposite direction of the motion (it removes energy from the system), hence the correct answer is e. -34 J.

User Shivendra Soni
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