Physics
posted by Tyler on .
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?

I think this is how you tackle this problem:
Fnet=TFg
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=550.0Nm*g
50.0kg*a=550.0N(50.0kg*9.8m/s^2)
55.0kg*a=550.0N490N
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