On a sky coaster (human pendulum) that reaches 20 meters from it’s equilibrium position, a man of 70 kg is able to reach a maximum speed of

Question

On a sky coaster (human pendulum) that reaches 20 meters from it’s equilibrium position, a man of 70 kg is able to reach a maximum speed of

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Khoii Minh 3 years 2021-09-05T08:17:23+00:00 2 Answers 2 views 0

Answers ( )

    0
    2021-09-05T08:19:06+00:00

    Answer:

    Explanation:

    Using the principle of conservation of energy, the potential energy is converted to kinetic energy, assuming any losses.

    Kinetic energy is given by ½mv²

    Potential energy is given by mgh

    Where m is the mass, v is the velocity, g is acceleration due to gravity and h is the height.

    Equating kinetic energy to be equal to potential energy then

    ½mv²=mgh

    V

    Making v the subject of the formula

    v=√(2gh)

    Substituting 9.81 m/s² for g and 20 m for h then

    v=√(2*9.81*20)=19.799 m/s

    Rounding off, v is approximately 20 m/s

    0
    2021-09-05T08:19:09+00:00

    Answer:

    19.8 m/s

    Explanation:

    During the motion of a pemdulum bob, it casually converts kinetic energy to potential energy and vice versa.

    A pendulum bob reaches its maximum speed at a position closest to its equilibrium position and has its lowest when it is farthest from the equilibrium position.

    The maximum speed of a pendulum bob based on the mass involved and the maximum displacement from the equilibrium position is obtained from

    Maximum kinetic energy = Maximum potential energy

    Maximum potential energy occurs at the farthest point from equilibrium, that is,

    P.E(max) = mgh

    Maximum kinetic energy = ½mv²

    ½mv² = mgh

    v = √2gh

    g = acceleration due to gravity = 9.8 m/s²

    h = farthest height from equilibrium position = 20 m

    v = √(2×9.8×20) = 19.8 m/s

    Hope this Helps!!!

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