Physics Work & Energy MDCAT 2018
PMDC Verified Question 83 of 108
A stone of mass 2.0 kg is dropped from a rest position 5.0m above the ground. What is its velocity at a height of 3.0m above the ground?
A
12.5m/s
B
9.3m/s
C
6.3m/s
D
16.0m/s
Tap any option to test your recall and reveal the step-by-step Propolis autopsy.

Propolis Cognitive Error Autopsy

Official Correct Choice:
Option C: 6.3m/s
Concept:

Using the Law of Conservation of Energy, the loss in potential energy equals the gain in kinetic energy. The velocity depends only on the distance it has fallen, not the mass.

Formula:

$$ v = \sqrt{2g\Delta h} $$

Solution:

  • Initial height = 5.0 m


  • Final height = 3.0 m


  • Distance fallen (\( \Delta h \)) = 5.0 - 3.0 = 2.0 m


  • \( v = \sqrt{2 \times 9.8 \times 2.0} = \sqrt{39.2} \)


  • Since \( 6^2 = 36 \) and \( 7^2 = 49 \), \( \sqrt{39.2} \) is slightly more than 6.


  • \( v \approx 6.26 \text{ m/s} \), which rounds to 6.3 m/s.


Why other options are incorrect:

Using the remaining height (3m) instead of the fallen distance gives \( \sqrt{2 \times 9.8 \times 3} \approx 7.7 \text{ m/s} \). Calculating velocity at the very bottom (h=5m) gives \( \sqrt{98} \approx 9.9 \text{ m/s} \).

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