Air resistance and magnitude

[brake4country]

If a 10 kg mass is in free fall (no air resistance), what force must be applied to slow the object down? (g = -10 m/s^2.) For this question, why is it 200 N and not 100 N? Wouldn't 100 N cancel out upward and downward forces and cancel them out?

 

[berkeman]

If a 10 kg mass is in free fall (no air resistance), what force must be applied to slow the object down? (g = -10 m/s^2.) For this question, why is it 200 N and not 100 N? Wouldn't 100 N cancel out upward and downward forces and cancel them out?

Welcome to the PF.

Can you state the exact question? That may make it easier to help you with this. What happens to the object's motion when you apply an equal force upward, so the sum of the forces is zero? If the object has an initial velocity downward before you apply the equalizing force, what happens to that velocity? What is the net acceleration of the object?

 

[brake4country]

Yes, the problem is as stated:

A 10 kg mass is in free fall with no air resistance. In order to slow the mass at a rate equal to the magnitude of g, an upward force must be applied with magnitude:

(a) 0 N
(b) 10 N
(c) 100 N
(d) 200 N

Thanks in advance to whomever helps me understand this!

 

[NascentOxygen]

What do you picture happening to the body's motion if you were to apply a force opposing its motion of 101 N?

 

[brake4country]

A net force of 1 N upward. Is this rationale correct? Wouldn't a force of 200 N blast this mass upward, instead of slowing it down?

 

[NascentOxygen]

A net force of 1 N upward. Is this rationale correct? Wouldn't a force of 200 N blast this mass upward, instead of slowing it down?

What are the equations of motion you will use to prove or disprove this?

Don't overlook the fact that before it can hurtle upwards (if in fact that were to be the outcome), its descent must be checked, slowed, and stopped, and then reversed. None of this will happen in an instant.

 

[brake4country]

The only equation I can think of relevant to this problem is F=ma. So the force moving upward to slow it down must be greater than 100 N, for which the only answer choice is 200 N?

 

[NascentOxygen]

We aren't interested in merely slowing it down, though. You need to address the requirement "to slow the mass at a rate equal to the magnitude of g".

 

[brake4country]

Ok, I misread the question. If the object is being slowed instead of being stopped, why is the answer still 200 N? Isn't that going to give a net force of 100 N upward? I am so confused on this question. I am thinking in terms of F=ma and vectors.

 

[NascentOxygen]

The examiner has chosen his words very carefully, leaving no room for ambiguity, even though there may be if it were viewed in the vernacular.

When the body is in free-fall, experiencing no air resistance, we know it gets faster at rate 9.8m/sec². The question stipulates a change so the body experiences an acceleration not of zero, but of -9.8m/sec². This means, if applied for an extended time, the force achieving this will not only slow the body, and stop it, but also reverse its direction of travel.

 

[Redbelly98]

Isn't that going to give a net force of 100 N upward?

Correct.

And therefore, the acceleration of that 10 kg object would be ___?

 

[brake4country]

Ok. So, a net force of 100 N upward will have no acceleration and constant velocity? Therefore, in order to change to slow it down since it is in constant velocity, another net force of 100 N would need to be applied (hence, an acceleration of 10 m/s in the upward motion). Is this correct?

 

[NascentOxygen]

Ok. So, a net force of 100 N upward will have no acceleration and constant velocity? Therefore, in order to change to slow it down since it is in constant velocity, another net force of 100 N would need to be applied (hence, an acceleration of 10 m/s in the upward motion). Is this correct?

http://imageshack.com/a/img29/6853/xn4n.gif but scratch out "net".

You are misusing the term "net", it means the sum of forces. Zero acceleration is the result of zero net force on the body. Weight counts as one force.