Question

Using a simple pulley/rope system, a crewman on an arctic expedition is trying to lower a 5.26 kg crate to the bottom of a steep ravine of height 28.6 meters. The 60.9 kg crewman is being careful to lower the crate at a constant speed of 1.50m/s. Unfortunately, when the crate reaches a point 15.9 meters above the ground, the crewman slips and 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.


_______m/s
At what speed will the crewman hit the Bottom of the ravine? (assume no air friction)
________m/s

Answer 1

Answer:

Explanation:

Mass of crate m = 5.26kg

Mass of crewman M = 60.9 kg

velocity of crate downwards u = 1.5 m/s

Height of ravine H = 28.6 m

Height of crate when crewman slips h = 15.9 m .

When crewman slips , Both M + m will have a common acceleration of

a = mg / ( M+m )

= (5.26 x 9.8 ) / ( 5.26 + 60.9 )

= 0.78 m / s²

We can calculate the final velocity of crate after going down by height 15.9 m from the following  formula

V² = u² + 2ah

= 1.5² + 2 x 0.78 x 15.9

V = 5.2 m/s

The crew will have both horizontal and vertical velocity.

Horizontal velocity will have acceleration of .78 m/s² with distance travelled  

is 15.9 m

v² = 0 + 2 x .78 x 15.9

v = 4.98 m /s

Vertical velocity will be created by free fall of 15.9 m

v² = 2 g h

2 x 9.8 x 15.9

v = 17.65 m/s

Resultant velocity

V(R)² = 17.65² + 4.98²

V(R) = 18.34 m/s