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

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

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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