Asked by Tyler
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?
Answers
Answered by
Devron
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
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
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.