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A 1.00-kilogram ball is dropped from the top of a building. just before striking the ground, the ball's speed is 12.0 meters per second. what was the ball's gravitational potential energy, relative to the ground, at the instant it was dropped? [neglect friction.]

Respuesta :

during the fall, the potential energy stored in the ball is converted into kinetic energy.
Thus,
PE = KE before hitting the ground
= 1/2 • mv^2
= 1/2 • 1 • 12^2
= 72J
abidemiokin

The ball's gravitational potential energy, relative to the ground, at the instant it was dropped is 72 Joules

  • During the fall, the potential energy of the ball is converted to kinetic energy (energy possessed by a body by virtue of ist motion).

The formula for calculating the kinetic energy of the ball is expressed as:

  • [tex]PE = KE = \frac{1}{2}mv^2[/tex]

  • m is the mass of the ball = 1.0kg
  • v is the ball's speed = 12m/s

Substitute into the formula:

[tex]GPE =\frac{1}{2}(1)(12)^2\\GPE=\frac{144}{2}\\GPE=72Joules[/tex]

Hence the ball's gravitational potential energy, relative to the ground, at the instant it was dropped is 72 Joules

Learn more on potential energy here: https://brainly.com/question/13584911

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