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 an

Question

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?

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Ngọc Hoa 1 month 2021-08-15T21:03:08+00:00 2 Answers 0 views 0

Answers ( )

    0
    2021-08-15T21:04:35+00:00

    Answer:

    T = 228.3N

    Explanation:

    We are given:

    w = 130N

     v = 5.5m/s^2

    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:

     T - mg = mv^2 / r ;

    Since we are asked to find tension T, we now have:

     T = mg + (mv^2)/r

    Therefore, substituting figures in the equation, we have:

     T = 130+(13 * 5.5^2)/4 ;

    T = 228.3N

    The tension in the rope is 228.3N

    0
    2021-08-15T21:04:43+00:00

    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

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