3 years ago

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

Physics

1 answer:

aleksandrvk [35]3 years ago

8 0

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

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

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