A 30-N stone is dropped from a height of 10 m and strikes the ground with a speed of 13 m/s. What average force of air friction acted on the stone as it fell?
A) 4.1 N
B) 2.9 N
C) 7.2 N
D) 1.2 N
E) 0.13 kN

Question

Quizplus · Accepted Answer

The average force of air friction can be calculated by first determining the work done by air friction and then using the work-energy principle. The kinetic energy (KE) when the stone hits the ground can be calculated using the formula KE = 0.5 * m * v^2, where m is the mass (in kg) and v is the velocity (in m/s). The potential energy (PE) lost by the stone can be calculated using PE = m * g * h, where g is the acceleration due to gravity (9.8 m/s^2) and h is the height (in meters). The work done by air friction is the difference between the initial potential energy and the final kinetic energy. The force of air friction can then be calculated by dividing the work done by air friction by the distance over which it acted (10 m in this case).First, convert the weight of the stone to mass: $$ F = m \cdot g \Rightarrow m = \frac{F}{g} = \frac{30\,N}{9.8\,m/s^2} \approx 3.06\,kg $$Calculate the kinetic energy at the bottom:$$ KE = 0.5 \cdot m \cdot v^2 = 0.5 \cdot 3.06\,kg \cdot (13\,m/s)^2 \approx 258.39\,J $$Calculate the potential energy at the top:$$ PE = m \cdot g \cdot h = 3.06\,kg \cdot 9.8\,m/s^2 \cdot 10\,m = 300\,J $$The work done by air friction is the difference between the potential energy at the top and the kinetic energy at the bottom:$$ W_{\text{friction}} = PE - KE = 300\,J - 258.39\,J \approx 41.61\,J $$The average force of air friction, assuming it acts over the entire 10 m drop, is:$$ F_{\text{friction}} = \frac{W_{\text{friction}}}{d} = \frac{41.61\,J}{10\,m} = 4.16\,N $$Rounding to one decimal place gives 4.1 N, which corresponds to choice A.

A 30-N Stone Is Dropped from a Height of 10