Physics Work & Energy DUHS 2023
PMDC Verified Question 33 of 108
Which of the following expression is constant for a freely falling body?
A
\( mgh + mv^2 \)
B
\( mgh = mv^2 \)
C
\( mgh + \frac{1}{2}mv^2 \)
D
\( mgh = -\frac{1}{2}mv^2 \)
Tap any option to test your recall and reveal the step-by-step Propolis autopsy.

Propolis Cognitive Error Autopsy

Official Correct Choice:
Option C: \( mgh + \frac{1}{2}mv^2 \)
Concept:

According to the Law of Conservation of Mechanical Energy, in the absence of air resistance, the total mechanical energy of a freely falling body remains constant at every point in its path.

Formula:

$$ E_{\text{total}} = P.E + K.E = \text{Constant} $$

Solution:

  • Potential Energy (P.E) = \( mgh \)


  • Kinetic Energy (K.E) = \( \frac{1}{2}mv^2 \)


  • Their sum, \( mgh + \frac{1}{2}mv^2 \), represents the total mechanical energy, which is a conserved constant.


Why other options are incorrect:

Option A misses the \( 1/2 \) coefficient for kinetic energy. Option B and D misstate the conservation law as an equality between instantaneous PE and KE, which is only true at exactly half the initial height.

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