Field biologist Leema is out gathering fungi in the rain forest when she stumbles into a 4.0 m deep pit lined with killer snakes. Nasteho Warrior Princess hears Leema's cries for help and dashes to the pit to save her. Nasteho lowers a vine down into the pit - the breaking strength of the vine is 8.00x10^2 N - and starts to pull Leema out. If Leema's weight is 602 N, what is the maximum acceleration Nasteho can use to save her? How long does it take to pull her out of danger?
d = 4.0 m F(vine) = 800 N F(weight) = - 602 N g = 9.81 m/s^2 \[F(weight) = mg\]so m = (602)/(9.81) = 61.4 kg The net force acting on Leema is: F(net) = F(vine) + F(weight) = 800 - 602 = 198 N \[\Sigma F = ma\]so a = (198)/(61.4) = 3.2 m/s^2 \[d = 1/2at^2\]so t^2 = 4/(0.5*3.2) = 4/1.6 = 2.5 s^2 thus, t = 1.58 s Answer: a = 3.2 m/s^2; t = 1.58 s
Conceptually, we can assume that Nasteho is pulling the vine with a force that is equal to the breaking strength of the vine. As a result, the vine exerts a positive (upwards) force, while gravity exerts a negative (downwards) force. The net force acting on Leema will be equal to the sum of these two forces. The net force could then be used to find Leema's acceleration.
so now do i need to fine leema's acceleratin?
find*
You know that the net force acting on Leema is 198 N. Using the formula F = ma, you can find Leema's acceleration. And you need to find Leema's acceleration in order to find the time required to pull her out of danger.
umm...ok so do i hav F=ma?
Yes, you use F = ma. See my calculations above.
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