BR Physics Chapter 1 Passage 4 Question 25

[getfat]

Hey guys, just needed a little clarification.

25.Not correcting for wind resistance when evaluating the results of the experiment would lead to a value for the gravitational force constant that is too:
A. large. The standard deviation in the raw data would be affected by the presence of wind resistance.
B. large. The standard deviation in the raw data would not be affected by the presence of wind resistance.
C. small. The standard deviation in the raw data would be affected by the presence of wind resistance.
D. small. The standard deviation in the raw data would not be affected by the presence of wind resistance.

The answer is D , and the explanation is:
Wind resistance opposes the direction of the motion (it always opposes the velocity of a moving object). This means that there is an upward force on the ball as it undergoes free fall, resulting in a reduced overall downward force. The time it would take for the ball to fall is increased because of wind resistance leading to lower calculated value for the acceleration, so the value of g would be underestimated if you failed to correct for wind resistance. This eliminates choices A and B. The effect of wind resistance would be uniform across the data, because it increases as velocity increases. This means that no variation in the standard deviation should be observed beyond that of human error, so rule out choice C in favor of choice D. The best answer is choice D.

What is confusing me about this question is that I would assume without calculating wind resistance, the gravitational force would be OVERESTIMATED because wind resistance opposes gravity. So it would be LARGE.

Thanks a lot in advance!

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[syoung]

y = .5gt^2 if no initial velosity (dropping a ball)

 

[BerkReviewTeach]

Think about the time it would take to fall in a vacuum versus with wind resistance. The upward drag force slows the object down and it takes longer to fall and doesn't reach as high of a speed at impact. if you just consider the units of a (m/s^2) then as time goes up, you have a large denominator and therefore a smaller a. You can also consider the definition where a = [delta]v/time. With a drag force, the [delta]v is smaller (it doesn't speed up as much) and the time to experience this reduced change in velocity is greater, so with [delta]v being small and t being bigger, the a must become lower.

To visualize, take it to extremes. Drop a golf ball in air and then drop it through water. There is also the buoyant force now at play, but the point being made by this visual is that when there is a force opposing gravity, a falling object will not speed up as readily (i.e., it has a reduced acceleration).