Question

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?

Answer 1

a. Vc = 5.06 m/s

b. Vp = 22.18 m/s

Step-by-step explanation:

The acceleration of the pulley-mass system is as follows:

a =
`(mg)/(m + M)`

Solving for acceleration, we get:

a =
`(6.17 *9.8)/(6.17 + 55.6)`

a = 0.97

So, for the part a:

Calculate the velocity of the crewman by using the following equation:

Vc =
`\sqrt{Vi^(2) + 2ay}`

Substituting the values into the equation, we get:

Vc =
`\sqrt{1.50^(2) + 2*0.97*13.2}`

Vc = 5.06 m/s

Now, for part b:

Calculate the final velocity of the pulley by using the following expression:

Vp =
`\sqrt{Vi^(2) +2gy }`

Just plugging in the values.

Vp =
`\sqrt{5.06^(2) +2*9.8*23.8 }`

Vp = 22.18 m/s