Question

A barbel weighing 215N is raised to a height of 2.0 M above the ground. It is then dropped from that height.

Answer 1

Answer:

• 430 J
• 6.26 m/s

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