Using a simply pulley/rope system, a crewman on an Arctic expedition is trying to lower a 6.17-kg crate to the bottom of a steep ravine of height 23.8 meters. The 55.6-kg crewman is walking along holding the rope, being careful to lower the crate at a constant speed of 1.50 m/s. Unfortunately, when the crate reaches a point 13.2 meters above the ground, the crewman steps on a slick patch of ice and slips. The crate immediately accelerates toward the ground, dragging the hapless crewman across the ice and toward the edge of the cliff.
______If we assume the ice is perfectly slick (that is, no friction between the crewman and the ice once he slips and falls down), at what speed will the crate hit the ground? Assume also that the rope is long enough to allow the crate to hit the ground before the crewman slides over the side of the cliff.
______At what speed will the crewman hit the bottom of the ravine? (Assume no air friction.)
Well, this sounds like quite the slippery situation! Let's break it down step by step, or should I say "slide by slide":
First, we need to find the acceleration of the crate as it falls. The crewman is lowering it at a constant speed of 1.50 m/s, which means the net force on the crate is zero. So, the gravitational force pulling the crate down must be equal to the force the crewman is applying upward.
The gravitational force on the crate can be calculated using F = m * g, where m is the mass of the crate (6.17 kg) and g is the acceleration due to gravity (approximately 9.8 m/s^2). So, the gravitational force on the crate is 6.17 kg * 9.8 m/s^2 = 60.386 N.
Now, let's move on to the crewman. When he slips, there's no friction between him and the ice, so all we have is the force of gravity acting on him. The crewman's mass is 55.6 kg, so the force of gravity on him is 55.6 kg * 9.8 m/s^2 = 544.88 N.
Since the crate is being accelerated downward and the crewman is being accelerated downward as well, we can treat it as a system where the crate and the crewman have the same acceleration.
Using Newton's second law (F = m * a), we can equate the gravitational forces on the crate and the crewman to find their acceleration:
60.386 N = (6.17 kg + 55.6 kg) * a
Now we can solve for a:
a = 60.386 N / 61.77 kg ≈ 0.977 m/s^2
Okay, now that we know the acceleration, we can find the final speed of the crate when it hits the ground. We can use the kinematic equation v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity (1.50 m/s), a is the acceleration, and s is the distance traveled (23.8 m - 13.2 m = 10.6 m).
Plugging in the values, we have:
v^2 = (1.50 m/s)^2 + 2 * 0.977 m/s^2 * 10.6 m
v^2 = 2.25 m^2/s^2 + 20.684 m^2/s^2
v^2 = 22.934 m^2/s^2
v ≈ √22.934 m^2/s^2
v ≈ 4.79 m/s
So, the crate will hit the ground with a speed of approximately 4.79 m/s.
Now, let's find out the speed at which the crewman hits the bottom of the ravine.
Since we know the acceleration (0.977 m/s^2) and the distance traveled (23.8 m - 13.2 m = 10.6 m), we can use the same kinematic equation to find the final speed of the crewman:
v^2 = (0 m/s)^2 + 2 * 0.977 m/s^2 * 10.6 m
v^2 = 0 m^2/s^2 + 20.684 m^2/s^2
v^2 = 20.684 m^2/s^2
v ≈ √20.684 m^2/s^2
v ≈ 4.55 m/s
So, the crewman will hit the bottom of the ravine with a speed of approximately 4.55 m/s.
I hope this icy adventure didn't leave you too chilled! Stay safe out there!
To solve this problem, we need to consider the conservation of energy.
Step 1: Calculate the potential energy of the crate at 13.2 meters above the ground using the formula:
Potential energy (PE) = mass (m) x gravitational acceleration (g) x height (h)
Given:
Mass of the crate (m) = 6.17 kg
Height above the ground (h) = 13.2 m
Gravitational acceleration (g) = 9.8 m/s^2
Potential energy (PE) = 6.17 kg x 9.8 m/s^2 x 13.2 m
Step 2: Calculate the kinetic energy of the crate when it hits the ground using the conservation of energy:
Kinetic energy (KE) = Potential energy (PE) at 13.2 meters above the ground
Step 3: Calculate the velocity (speed) of the crate when it hits the ground using the formula:
Kinetic energy (KE) = (1/2) x mass (m) x velocity^2
Given:
Mass of the crate (m) = 6.17 kg
Velocity of the crate (v) = unknown
Now we can solve for the velocity (v) using the formula:
v = sqrt(2 x KE / m)
Step 4: Calculate the velocity of the crewman when he hits the bottom of the ravine using the conservation of energy:
Since there is no air friction, the total mechanical energy (kinetic energy + potential energy) of the crewman will remain constant.
Potential energy (PE) of the crewman = Potential energy (PE) of the crate
Kinetic energy (KE) of the crewman = Kinetic energy (KE) of the crate
Given:
Mass of the crewman (m') = 55.6 kg
Velocity of the crewman (v') = unknown
Now we can solve for the velocity (v') using the formula:
v' = sqrt(2 x KE / m')
Let's plug in the values and calculate the results.
Step 1: Calculate the potential energy of the crate at 13.2 meters above the ground:
PE = 6.17 kg x 9.8 m/s^2 x 13.2 m
PE = 1087.224 Joules
Step 2: Calculate the kinetic energy of the crate when it hits the ground:
KE = 1087.224 Joules
Step 3: Calculate the velocity (speed) of the crate when it hits the ground:
v = sqrt(2 x 1087.224 Joules / 6.17 kg)
v ≈ 15.7 m/s
Therefore, the crate will hit the ground with a speed of approximately 15.7 m/s.
Step 4: Calculate the velocity of the crewman when he hits the bottom of the ravine:
v' = sqrt(2 x 1087.224 Joules / 55.6 kg)
v' ≈ 6.87 m/s
Therefore, the crewman will hit the bottom of the ravine with a speed of approximately 6.87 m/s.
To find the speed at which the crate hits the ground, we can use the principle of conservation of energy. At the initial position, the only form of energy the crate has is potential energy, given by mgh, where m is the mass of the crate (6.17 kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the height of the ravine (23.8 m). At the final position, when the crate hits the ground, all the potential energy is converted into kinetic energy, given by (1/2)mv^2, where v is the final velocity of the crate.
Setting these two energies equal and solving for v, we have:
mgh = (1/2)mv^2
Canceling out the mass of the crate and rearranging the equation, we get:
v^2 = 2gh
Plugging in the values, we have:
v^2 = 2 * 9.8 m/s^2 * 23.8 m
Simplifying, we find:
v^2 = 446.92
Taking the square root of both sides, we have:
v = √446.92
So, the speed at which the crate hits the ground is approximately 21.14 m/s.
To find the speed at which the crewman hits the bottom of the ravine, we can again use the principle of conservation of energy. In this case, the crewman loses potential energy as he slides down the ravine, and this is converted into kinetic energy.
At the initial position, the crewman is at the same height as the crate (13.2 m), so his potential energy is given by mgh, where m is the mass of the crewman (55.6 kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the initial height (13.2 m). At the final position, when the crewman hits the bottom of the ravine, all the potential energy is converted into kinetic energy.
Setting these two energies equal and solving for v, we have:
mgh = (1/2)mv^2
Canceling out the mass of the crewman and rearranging the equation, we get:
v^2 = 2gh
Plugging in the values, we have:
v^2 = 2 * 9.8 m/s^2 * 13.2 m
Simplifying, we find:
v^2 = 258.72
Taking the square root of both sides, we have:
v = √258.72
So, the speed at which the crewman hits the bottom of the ravine is approximately 16.1 m/s.