Final answer:
The work done by the drag force on an object falling with terminal velocity is equal to the negative of the work done by gravity, which can be calculated using the mass of the object, the difference in height, and the acceleration due to gravity.
Step-by-step explanation:
The student is asking how much work the drag force did on an object with a mass of 0.126 kg as it fell from a height of 15.2 m to 7.9 m, reaching terminal velocity v in the process. To calculate this, we use the work-energy principle which states that the work done by the drag force, plus the work done by gravity, is equal to the change in kinetic energy of the object.
Since the object is falling with terminal velocity, its speed does not change, meaning that the change in kinetic energy is zero. Therefore, the work done by the drag force is equal to the negative of the work done by gravity. The work done by gravity (Wg) is the product of the gravitational force (weight) and the displacement in the direction of the force, which is equal to m × g × (hf - hi), where m is the mass, g is the acceleration due to gravity (9.81 m/s2), hf is the final height, and hi is the initial height.
Plugging in the values gives us:
- m = 0.126 kg
- hf = 7.9 m
- hi = 15.2 m
We can calculate Wg = 0.126 kg × 9.81 m/s2 × (7.9 m - 15.2 m). Doing the math, Wg = -0.126 kg × 9.81 m/s2 × (-7.3 m) = 0.126 kg × 9.81 m/s2 × 7.3 m
After simplification, Wg is positive, which means that the work done by the drag force is negative this value, as it acts in the opposite direction to gravity.