Understanding Air Resistance: Explaining the Math Behind Reduced Force

AI Thread Summary
When air resistance on an object is reduced by a factor of four, the net force acting on the object becomes three-quarters of the original force. Initially, at constant velocity, the force of air resistance equals the gravitational force, balancing each other out. With the reduction in air resistance, the remaining force is calculated as 3/4 of the original force. This change leads to a net force that results in acceleration, as the object is no longer in equilibrium. Understanding this relationship between force and acceleration is crucial in grasping the effects of reduced air resistance.
brake4country
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Can someone please explain the math of this problem?

-If F is the force of air resistance on an object with mass m moving at a constant velocity, which of the following best describes the acceleration of the object when the force of air resistance is reduced by a factor of 4?
(a) F/m
(b) 1/2 F/m
(c) 1/4 F/m
(d) 3/4 F/m

The answer is d but I do not understand how to get it. Thanks in advance!
 
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If it is moving at a constant velocity, what force is required in order to maintain that velocity in the initial case (with full air resistance)?

If the Air resistance is reduced by a factor of four, but the same force is assumed to be acting on the object, how does the object react?
 
I am assuming that when velocity is constant, then the F(air) must cancel out the F of (mg). Therefore, when I drew this out on paper, I made the F(mg) component longer and the F(air) shorter since F(air) is reduced.

How do the fraction come into play? How is the answer 3/4 F/m? Thanks in advance
 
brake4country said:
I am assuming that when velocity is constant, then the F(air) must cancel out the F of (mg). Therefore, when I drew this out on paper, I made the F(mg) component longer and the F(air) shorter since F(air) is reduced.

How do the fraction come into play? How is the answer 3/4 F/m? Thanks in advance

What is the net force remaining when you reduce the air resistance by a factor of 4?
 
I get it. Reducing by a factor of 4 requires one to think in terms of fractions (1/4). Therefore, the net F remaining is 3/4. Thanks again!
 

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