Concept:Use the Conservation of Energy with non-conservative forces: The initial potential energy is converted into kinetic energy minus the energy lost to air resistance.
Formula:$$ P.E = K.E + W_{\text{friction}} \implies mgh = \frac{1}{2}mv^2 + W_f $$
Solution:- Initial P.E = \( 2 \times 10 \times 8 = 160 \text{ J} \) (Using \( g \approx 10 \) for simplicity).
- Energy lost = 50 J.
- Remaining energy for K.E = \( 160 - 50 = 110 \text{ J} \).
- Set K.E equal to 110: \( \frac{1}{2}(2)v^2 = 110 \).
- \( v^2 = 110 \implies v \approx 10.48 \text{ m/s} \).
- If we strictly use \( g = 9.8 \): \( P.E = 2 \times 9.8 \times 8 = 156.8 \text{ J} \). K.E = \( 156.8 - 50 = 106.8 \text{ J} \). \( v = \sqrt{106.8} \approx 10.3 \text{ m/s} \).
- In either case, 10 m/sec is by far the closest option.
Why other options are incorrect:Options B, C, and D are wildly large and physically impossible for an 8m drop even in a vacuum (max speed in vacuum is ~12.5 m/s).
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