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A rock of mass 200kg is dropped from a height of 200m. What is its potential energy at: a) 0 seconds b) Just before hitting the ground?

Physics

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Answers

2 parts
a

392000 J\text{392000 J}

b

0 J\text{0 J}

Step-by-step solution

4 steps

Here's how to calculate the potential energy:

The formula for gravitational potential energy (GPE) is:

GPE=mghGPE = mgh

where:

  • mm is the mass of the object
  • gg is the acceleration due to gravity (approximately 9.8 m/s29.8 \text{ m/s}^2)
  • hh is the height of the object above a reference point

Given:
Mass (mm) = 200 kg200 \text{ kg}
Initial height (h0h_0) = 200 m200 \text{ m}
Acceleration due to gravity (gg) = 9.8 m/s29.8 \text{ m/s}^2

Part a· 2 steps

Potential energy at 0 seconds

  1. Determine the height at 0 seconds.

    At 0 seconds, the rock has just been dropped, so it is still at its initial height.

    h=200 mh = 200 \text{ m}
  2. Calculate the potential energy using the formula.

    GPE=mghGPE = mgh GPE=(200 kg)×(9.8 m/s2)×(200 m)GPE = (200 \text{ kg}) \times (9.8 \text{ m/s}^2) \times (200 \text{ m}) GPE=392000 JGPE = 392000 \text{ J}

    The potential energy at 0 seconds is 392000 J\boxed{\text{392000 J}}.

Part b

0 J\text{0 J}

Potential energy just before hitting the ground

  1. Determine the height just before hitting the ground.

    Just before hitting the ground, the height (hh) is approximately 0 m0 \text{ m}.

    h=0 mh = 0 \text{ m}
  2. Calculate the potential energy using the formula.

    GPE=mghGPE = mgh GPE=(200 kg)×(9.8 m/s2)×(0 m)GPE = (200 \text{ kg}) \times (9.8 \text{ m/s}^2) \times (0 \text{ m}) GPE=0 JGPE = 0 \text{ J}

    The potential energy just before hitting the ground is 0 J\boxed{\text{0 J}}.

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