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A 2.32-kg bucket of water is suspended from a string and swung in a vertical circle. The radius of the circle is 0.945 meters. The speed at the top of the circle is 3.92 m/s. Determine the (3) acceleration (in mis2), (4) net force (in N) and the magnitudes of all individual forces (1 and 2) when the bucket is at the top of the circle. Use g= 9.80 m/s2​

User Urbanmojo
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1 Answer

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Final answer:

The acceleration at the top of the circle is 16.132 m/s^2, the net force is 37.429 N, and the magnitudes of the tension force and gravitational force are both 37.429 N and 22.736 N, respectively.

Step-by-step explanation:

When the bucket is at the top of the circle, it experiences centripetal acceleration directed towards the center of the circle. This acceleration can be calculated using the formula:

a = v^2 / r

where a is the acceleration, v is the speed, and r is the radius of the circle. Plugging in the given values, we get:

a = (3.92 m/s)^2 / 0.945 m = 16.132 m/s^2

To find the net force, we can use the equation:

F_net = m * a

where F_net is the net force, m is the mass, and a is the acceleration. Plugging in the given values, we get:

F_net = 2.32 kg * 16.132 m/s^2 = 37.429 N

At the top of the circle, there are two forces acting on the bucket: the tension force, T, directed towards the center of the circle, and the gravitational force, mg, directed downwards. The tension force provides the centripetal force required for circular motion. The magnitude of the tension force is equal to the net force:

T = F_net = 37.429 N

The magnitude of the gravitational force is given by:

mg = (2.32 kg) * (9.80 m/s^2) = 22.736 N

User Everyman
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