Question
Illinois Jones is being pulled from a snake pit with a rope that breaks if the tension in it exceeds 755 .
If Illinois Jones has a mass of 70.0 and the snake pit is 3.90 deep, what is the minimum time that is required to pull our intrepid explorer from the pit?
Can anyone help?
Confused
If Illinois Jones has a mass of 70.0 and the snake pit is 3.90 deep, what is the minimum time that is required to pull our intrepid explorer from the pit?
Can anyone help?
Confused
Answers
MathMate
It is a very nice problem, except that the question lacks units. I will assume metric units, which will make g = 9.8 m/s/s.
I restate the problem as follows:
"Illinois Jones is being pulled from a snake pit with a rope that breaks if the tension in it exceeds F = 755 N.
If Illinois Jones has a mass m of 70.0 kg and the snake pit is h=3.90 m deep, what is the minimum time that is required to pull our intrepid explorer from the pit? "
The force on the rope just to suspend Jones in the air would be mg=686N.
Reserve capacity for acceleration (assume the rope has no mass) is therefore 755-686=69N
Mass, m = 70 kg
Using F=ma
Acceleration, a = 69/70 kg = 0.986 m/s/s
Assuming the lift starts from initial velocity v0=0 m/s,
the time t (seconds) required to lift him through h=3.9 m is given by
h = v0.t + (1/2)at²
in which all variables are known except t.
So substitute all the known values and solve the resulting equation for t (in seconds). Do not retain the negative value of t. I believe it is under 3 seconds.
I restate the problem as follows:
"Illinois Jones is being pulled from a snake pit with a rope that breaks if the tension in it exceeds F = 755 N.
If Illinois Jones has a mass m of 70.0 kg and the snake pit is h=3.90 m deep, what is the minimum time that is required to pull our intrepid explorer from the pit? "
The force on the rope just to suspend Jones in the air would be mg=686N.
Reserve capacity for acceleration (assume the rope has no mass) is therefore 755-686=69N
Mass, m = 70 kg
Using F=ma
Acceleration, a = 69/70 kg = 0.986 m/s/s
Assuming the lift starts from initial velocity v0=0 m/s,
the time t (seconds) required to lift him through h=3.9 m is given by
h = v0.t + (1/2)at²
in which all variables are known except t.
So substitute all the known values and solve the resulting equation for t (in seconds). Do not retain the negative value of t. I believe it is under 3 seconds.
Cole
More like Junior year of high school physics, this identical problem is in my textbook from last year.
nick
man 14 years later i have this same problem....
Mackenzie
Me too. I am ready to be through with physics.