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21 November, 16:36

A 2.20kg pendulum starts from a height of 5.00m. It swings back and forth through one whole oscillation but only returns to a maximum height of 4.75m. How much negative work was done on the pendulum during the first entire oscillation?

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Answers (2)
  1. 21 November, 16:48
    0
    -5.396Joules

    Explanation:

    Work is said to be done when a force causes a body to move through a distance and acts in the direction of the force.

    Work done = Force * Distance

    Given force = mg

    Force = 2.20*9.81

    Force applied on the string = 21.58N

    Distance will be the difference in height = 4.75-5.00

    = 0.25m

    Negative work done on the pendulum during the first entirely oscillation = 21.58*0.25

    = - 5.396Joules
  2. 21 November, 18:20
    0
    Negative workdone = 5.39J

    Explanation:

    Mass (m) = 2.20kg

    h₁ = 5.0m

    h₂ = 4.75m

    g = 9.8m/s²

    Potential energy (PE) = mgh

    M = mass

    g = acceleration due to gravity

    h = height

    Work done = force * distance

    Force = mass * acceleration (a or g)

    Work done = mg * s

    P. E = mgs = mgh

    Negative potential energy of the pendulum = Mgh₂ - Mgh₁

    P. E = mg (h₂-h₁)

    P. E = 2.20 * 9.80 * (5.0 - 4.75)

    P. E = 21.56 * 0.75

    P. E = 5.39J

    The negative potential work done = 5.39J
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