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.)
To determine the speed at which the crate hits the ground and the crewman hits the bottom of the ravine, we can apply the principles of conservation of energy and Newton's second law.
Let's start by analyzing the situation when the crewman slips and falls.
First, let's find the potential energy of the crate at a height of 13.2 meters above the ground. The potential energy (PE) of an object at a certain height is given by the formula: PE = m * g * h, where m is the mass, g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height.
PE_crate = 6.17 kg * 9.8 m/s^2 * 13.2 m
PE_crate = 806.532 J
When the crewman slips, the crewman and the crate form a system in which their combined mass moves together. Therefore, the total potential energy of the system at this point is:
PE_system = PE_crate + PE_crewman
PE_system = 806.532 J + (55.6 kg * 9.8 m/s^2 * 13.2 m)
PE_system = 806.532 J + 8546.656 J
PE_system = 9353.188 J
Next, let's find the kinetic energy of the system at the bottom of the ravine. At this point, all the potential energy is converted to kinetic energy since there is no air friction.
KE_system = PE_system
According to the conservation of energy principle, the total energy of a closed system remains constant. Therefore, the potential energy at the start (13.2 meters above the ground) is equal to the kinetic energy at the bottom of the ravine.
9353.188 J = (6.17 kg + 55.6 kg) * v^2 / 2
Solving for v (the velocity), we find:
v^2 = 9353.188 J * 2 / 61.77 kg
v^2 = 302.4512
v = √(302.4512)
v ≈ 17.40 m/s
So, the crate will hit the ground with a speed of approximately 17.40 m/s.
To find the speed at which the crewman hits the bottom of the ravine, we can use Newton's second law, which states that force is equal to mass multiplied by acceleration (F = m * a).
Since there is no air friction to slow down the crewman's acceleration, the acceleration experienced by the crewman while falling will be approximately equal to the acceleration due to gravity (g ≈ 9.8 m/s^2).
Using the same principle, we can find the force acting on the crewman:
Force = mass_crewman * acceleration_due_to_gravity
Force = 55.6 kg * 9.8 m/s^2
Force = 545.68 N
Now, using the equation F = m * a, we can solve for the acceleration experienced by the crewman:
Force = mass_crewman * acceleration
545.68 N = 55.6 kg * acceleration
acceleration = 545.68 N / 55.6 kg
acceleration ≈ 9.801 m/s^2
Finally, we can use the equation of motion v^2 = u^2 + 2 * a * s, where v is the final velocity, u is the initial velocity (which is zero in this case), a is the acceleration, and s is the distance traveled.
v^2 = 0 + 2 * 9.801 m/s^2 * 23.8 m
v^2 ≈ 466.7936
v = √(466.7936)
v ≈ 21.60 m/s
Therefore, the crewman will hit the bottom of the ravine with a speed of approximately 21.60 m/s.