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?
2 answers:
Answer: 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
Answer:
T = 228.3N
Explanation:
We are given:
w = 130N
r = 4.0m
We take g = 10
Therefore:
w = m×g
130 = m × 10
m = 130/10
m = 10
At the lowest point,we use the equation:
;
Since we are asked to find tension T, we now have:
Therefore, substituting figures in the equation, we have:
;
T = 228.3N
The tension in the rope is 228.3N
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