Posted by **Tyler ** on Wednesday, May 14, 2014 at 9:48pm.

A person of mass 50.0 kg is at the bottom of a cave. The cave has a depth of 32.5 meters and the person is to be pulled out of the cave vertically by a rope, starting from rest. What is the shortest amount of time that this could take if the rope can withstand a maximum tension of 550.0 N without breaking?

- Physics -
**Devron**, Wednesday, May 14, 2014 at 10:09pm
I think this is how you tackle this problem:

Fnet=T-Fg

Where

T=550.N

Fg=m*g

Fnet=m*a

m=50.0kg

g=9.8m/s^2

and

a=?

Solve for a:

m*a=550.0N-m*g

50.0kg*a=550.0N-(50.0kg*9.8m/s^2)

55.0kg*a=550.0N-490N

55.0kg*a=60N

a=60N/55.0kg

a=1.091m/s^2

You know acceleration, so solve for time (t) using the following kinematic equation:

d=Vit+1/2at^2

where

d=32.5m

Vi=0m/s^2

a=1.091m/s^2

t=?

Solve for t:

32.5=0+1/2(1.091m/s^2)t^2

t=sqrt*(65m/1.091m/s^2)

t=7.72s

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