A 5.0-kg pumpkin is dropped from the top of the RCA Building in New York City.If it falls through a distance of 380 m,and splats on the sidewalk with a speed of 52.0 m/s,what is the work done by air resistance during its fall?

Question

Quizplus · Accepted Answer

The work done by air resistance can be found by calculating the difference between the gravitational potential energy lost and the kinetic energy gained by the pumpkin. The gravitational potential energy (GPE) lost is given by $mgh = (5.0\, \text{kg})(9.8\, \text{m/s}^2)(380\, \text{m}) = 18620\, \text{J}$ or 18.62 kJ. The kinetic energy (KE) gained by the pumpkin when it hits the ground is given by $\frac{1}{2}mv^2 = \frac{1}{2}(5.0\, \text{kg})(52.0\, \text{m/s})^2 = 6760\, \text{J}$ or 6.76 kJ. The work done by air resistance is the difference: $18.62\, \text{kJ} - 6.76\, \text{kJ} = 11.86\, \text{kJ}$, which rounds to approximately 12 kJ. Since air resistance does work against the motion, it is considered negative work, making the correct answer -12 kJ.

A 50-Kg Pumpkin Is Dropped from the Top of the RCA