Einswine (m = 20.0 kg) is trapped at the bottom of a hole. Luckily, Physics Girl comes to save him! She drops a rope down the hole. Einswine grabs onto the rope and Physics Girl starts pulling him vertically upwards. The maximum tension that the rope can withstand is 220.0 N. What is the shortest time, starting from rest, in which Einswine can be lifted out of the hole if the hole is 10.0 m deep?
To find the shortest time in which Einswine can be lifted out of the hole, we can use the concept of work and energy.
First, let's calculate the work done by Physics Girl to lift Einswine out of the hole. The work done is equal to the change in potential energy.
The potential energy of Einswine at the bottom of the hole is given by the formula:
PE = m * g * h
where
PE is the potential energy,
m is the mass of Einswine (20.0 kg),
g is the acceleration due to gravity (9.8 m/s^2),
h is the depth of the hole (10.0 m).
Substituting the given values:
PE = 20.0 kg * 9.8 m/s^2 * 10.0 m
= 1960 J
The maximum tension that the rope can withstand is the force required to lift Einswine out of the hole. This tension force is equal to the weight of Einswine plus any additional force required to accelerate him.
T = m * g + F
where
T is the tension force (220.0 N),
m is the mass of Einswine (20.0 kg),
g is the acceleration due to gravity (9.8 m/s^2),
F is the additional force required to accelerate Einswine.
Since Einswine starts from rest and reaches some final velocity, we can apply Newton's second law of motion:
F = m * a
where
F is the force (additional) required to accelerate Einswine,
m is the mass of Einswine (20.0 kg),
a is the acceleration of Einswine.
In this case, we want to find the shortest time, so the acceleration needs to be maximum. The maximum acceleration can be achieved if the tension force is equal to the maximum tension the rope can withstand. So we have:
220.0 N = (20.0 kg * 9.8 m/s^2) + (20.0 kg * a)
220.0 N = 196.0 N + (20.0 kg * a)
24.0 N = 20.0 kg * a
a = 1.2 m/s^2
Now, we can use the kinematic equation to calculate the shortest time:
v = u + (a * t)
where
v is the final velocity (0 m/s as Einswine comes to rest at the top of the hole),
u is the initial velocity (0 m/s as he starts from rest),
a is the acceleration (1.2 m/s^2),
t is the time taken.
Substituting the values:
0 m/s = 0 m/s + (1.2 m/s^2 * t)
Simplifying the equation:
1.2 m/s^2 * t = 0
t = 0 / 1.2 m/s^2
t = 0 s
Therefore, the shortest time it took to lift Einswine out of the hole is 0 seconds.