Physics Work & Energy MDCAT 2017
PMDC Verified Question 90 of 108
If mass 'm' is dropped from height 'h' vertically, f is the force of friction during downward motion and 'v' is the velocity at bottom, following equation will be hold:
A
\( \frac{1}{2} mv^2 = mgh + fh \)
B
\( fh = mgh + \frac{1}{2} mv^2 \)
C
\( mgh = \frac{1}{2} mv^2 - fh \)
D
\( mgh = \frac{1}{2} mv^2 + fh \)
Tap any option to test your recall and reveal the step-by-step Propolis autopsy.

Propolis Cognitive Error Autopsy

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

According to the Conservation of Energy in the presence of non-conservative forces, the initial potential energy is converted into kinetic energy plus the work done against friction.

Formula:

$$ E_{\text{initial}} = E_{\text{final}} + W_{\text{friction}} $$

Solution:

  • Initial energy at height \( h \) is entirely Potential Energy: \( P.E = mgh \)


  • Final energy at the bottom is Kinetic Energy: \( K.E = \frac{1}{2}mv^2 \)


  • Work done against air friction over distance \( h \): \( W = fh \)


  • Equating them: \( mgh = \frac{1}{2}mv^2 + fh \)


Why other options are incorrect:

Option A implies energy is created. Option B implies friction work equals total energy. Option C incorrectly subtracts friction work instead of adding it to the final state energies.

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