A barbel weighing 215N is raised to a height of 2.0 M above the ground. It is then dropped from that height. A. What is the kinetic energy of the barbell as it hits the floor?
B. What is the velocity of the barbell as it hits the floor?

Explanation:

A. The kinetic energy is the same as the initial potential energy:

  PE = mgh = (215 N)(2.0 M) = 430 J

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B. The velocity achieved by falling from a height h is given by ...

  v = √(2gh)

  v =  √(2·9.8 m/s^2·2 m) = √(39.2 m^2/s^2)

  v ≈ 6.26 m/s

all of the answers are correct.

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