A $7.5 \mathrm{~kg}$ bowling ball is dropped from the top of a $25 \mathrm{~m}$ tall building. Ignoring air resistance, just before the bowling ball hits the sidewalk below, its kinetic energy is equivalent to how many food Calories? (one food Calorie $=1$ kilocalorie in SI units)

Question

Quizplus · Accepted Answer

The kinetic energy just before hitting the sidewalk can be found using the formula for gravitational potential energy at the start (since all of it converts to kinetic energy by the end), which is $PE = mgh$, where $m = 7.5 \, \text{kg}$, $g = 9.8 \, \text{m/s}^2$, and $h = 25 \, \text{m}$. So, $PE = 7.5 \times 9.8 \times 25 = 1837.5 \, \text{J}$. Since $1 \, \text{food Calorie} = 4184 \, \text{J}$, the energy in food Calories is $1837.5 / 4184 \approx 0.44$.

A 7.5 kg7.5 \mathrm{~kg}7.5 kg Bowling Ball Is Dropped from the Top of A

A $7.5 \mathrm{~kg}$ Bowling Ball Is Dropped from the Top of A