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A bucket of water (m=2.0 kg) is attached to a 0.80-m long string and spun in a vertical circle. The speed of the water at the bottom of the circular path is 6.0 m/s. (a) What is the force of gravity? (b) What is the net force? (c) What is the force of tension? (d) What is the force of tension at the top of the circle?

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Final answer:

The force of gravity is 19.6 N, the net force is 45 N, the force of tension is 25.4 N at the bottom and 64.6 N at the top.

Step-by-step explanation:

To answer this question, we need to consider the forces acting on the bucket of water as it spins in a vertical circle.

(a) The force of gravity or weight can be calculated using the formula F = mg, where m is the mass of the bucket (2.0 kg) and g is the acceleration due to gravity (9.8 m/s^2). So the force of gravity is F = 2.0 kg * 9.8 m/s^2 = 19.6 N.

(b) The net force is the difference between the force of gravity and the force of tension. At the bottom of the circular path, the net force is equal to the centripetal force, which is given by F = m * v^2 / r, where m is the mass of the bucket (2.0 kg), v is the speed of the water (6.0 m/s), and r is the radius of the circular path (0.80 m). So the net force is F = 2.0 kg * (6.0 m/s)^2 / 0.80 m = 45 N.

(c) The force of tension can be found by subtracting the force of gravity from the net force. So the force of tension is T = net force - force of gravity = 45 N - 19.6 N = 25.4 N.

(d) At the top of the circle, the force of tension is the sum of the force of gravity and the net force. So the force of tension is T = net force + force of gravity = 45 N + 19.6 N = 64.6 N.

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