A heavy ball with a weight of 130 N is hung from the ceiling of a lecture hall on a 4.0-{\rm m}-long rope. The ball is pulled to one side and released to swing as a pendulum, reaching a speed of 5.5 m/s as it passes through the lowest point. What is the tension in the rope at that point?

The tension on the rope as the ball passes through the lowest point is 230.35 N.

The given parameters;

The tension on the rope as the ball passes through the lowest point of the circular motion is calculated as follows;

where;

The mass of the ball is calculated by applying Newton's law;

W = mg

The tension on the rope is calculated as;

Thus, the tension on the rope as the ball passes through the lowest point is 230.35 N.

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T = 228.3N

Explanation: The weight of the ball will be equal to the product of its mass and acceleration due to gravity.

weight, w = mg 

Then we can get the mass.

130 = mX10 

m = 13 kg 

Now let's consider all the forces at the lowest point.

The sum of the forces will be equal to centripetal force because it swings.

Therefore:

T - mg = mv^2 / r

T = 130 + (13 X 5.5^2)/4.0 = 130 + 98.3 = 228.3 N

the right answer is 0.3 meters per second