Falling flowerpotstuck on simple part

Rockdog

A 3.2 kg flowerpot drops from a tall building. The initial speed of the pot is zero, and you may neglect air resistance.

a) What is the magnitude of the force acting on the pot while it is in the air 1.5 s after it begins to fall?

b) After the pot has fallen 28 m, what is its speed?

c) After the pot has fallen 28 m, it enters a viscous liquid, which brings it to rest over a distance of 1.5 m. Assuming constant deceleration over this distance, what is the magnitude of this deceleration?

d) What is the force exerted on the liquid by the pot?

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Okay, I got the answers for the first three, which are
1) 31.36 N
2) 23.4 m/s
3) 182 m/s/s

If you really want to know how I did it, I can provide that in a later post, but thats not the point here.

I'm stuck on the last question.
I know I have to use this equation; F=M*a (in the y direction)
Well, I was thinking, the object has a downward force of 31.36 N, so when it hits the liquid, shouldn't the liquid push back up 31.36N? Of course, my answer doesn't match the computer, so insight appreciated.

*31.36N achieved by 3.2kg*9.8m/s/s

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Hurkyl

Staff Emeritus
Gold Member
The object does not have a downward force of 31.36 N... gravity is applying a 31.36 N downward force to the object. (I'm not sure if you know the distinction, so I'm stating it to be on the safe side)

You know the acceleration experienced by the object. You know the mass of the object. Thus, you know the net force acting on the object! (If the net force was 0, your reasoning would be correct in spirit... but the object is decelerating so the net force is not 0!)

There are two contributions to the net force; the 31.36 N downward force supplied by gravity, and the unknown force supplied by the liquid, and these have to add up to the net force.

Once you know the force with which the liquid acts upon the pot, then the negative of that is the force with which the pot acts upon the liquid.

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