Compare the kinetic energy (=(1)/(2) m v^2) of a 1 -gram piece of ice (about half the mass of a dime) entering Earth's atmosphere at a speed of 50 km / s to that of a 2 -metric-ton SUV (mass =2 × 10^3kg speeding down the highway at 90 km / h.

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Md Ariful Islam · Answer 1 · Apr 06, 2024 06:29:39

Kinetic Energy Comparison

We'll use the formula for kinetic energy:

\[ KE = \frac{1}{2} m v^2 \]

Given that the mass of the ice ($ m $) is 1 gram ($ 0.001 $ kg) and the velocity ($ v $) is 50 km/s, we have:

\[ KE_{\text{ice}} = \frac{1}{2} \times 0.001 \times (50)^2 \]

Using the same formula, but with the mass of the SUV ($ m $) as 2 metric tons ($ 2 \times 10^3 $ kg) and the velocity ($ v $) as 90 km/h, we have:

\[ KE_{\text{SUV}} = \frac{1}{2} \times 2 \times 10^3 \times \left(\frac{90}{3.6}\right)^2 \]