Using a simply pulley/rope system, a crewman on an Arctic expedition is trying t
ID: 1562542 • Letter: U
Question
Using a simply pulley/rope system, a crewman on an Arctic expedition is trying to lower a 5.52-kg crate to the bottom of a steep ravine of height frictionless 23.4 meters. The 50.3-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 13.0 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 figure not to scale 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.)Explanation / Answer
vi = 1.5m/sec
5.52g - T = 5.52a -----------1
T = 50.3a -----------2
eqn 1+2
5.52g = (55.82)a
=>a = 5.52*9.8/55.82
=>a = 0.969m/s2
h = 13 m
vy^2 - uy^2 = 2as
vy^2 - 1.5^2 = 2*0.969*13
vy^2 = 2.25 + 2*0.969*13
=> vy = 5.24 m/sec velocity at the time before time hitting ground
vy - uy=at
5.24 - 1.5 = 0.969t
=>t = 3.86 sec
assume the man reaches the edge of cliff when vy of block is 5.24 m/sec
uman = 1.5m/sec
a = 0.969m/s2
so Vman = uman + at
= 1.5 + 0.969*3.86
= 5.24 m/sec
so Vman x = 5.24 m/sec
Vmany inital =0
vman final ^2 - uman y initial ^2 = 2gh
Here h = 23.4 m
vman y final^2 = 0+2*9.8*23.4
vman y final = 21.42 m/sec
velocity of man at the bottom of cliff is 5.24 i - 21.42 j
magnitude of man's velocity = 22.05 m/sec
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.