Prof Lippert enters an elevator on the 3rd floor of the Metcalfe building to go

Question

Prof Lippert enters an elevator on the 3rd floor of the Metcalfe building to go up to his office on the 11th floor. The elevator starts 8 m above the ground and begins accelerating upward at 5.0 m/s^2 for 1.0 s. Then, a disgruntled student cuts the elevator cables, causing the elevator to fall freely back toward the ground. What is the velocity of the elevator at the moment the cables arc cut? How high off the ground is the elevator when the cables are cut? What is the velocity of the elevator when it hits the ground? How long after the cables arc cut does the elevator hit the ground?

Explanation / Answer

part a:

initial speed=0 m/s

acceleration=5 m/s^2

time taken=1 second

then veloicty at the time cable was cut=initial velocity+acceleration*time

=0+5*1=5 m/s

part b:

distance covered by the elevator before cable cutting=

initial velocity*time+0.5*acceleration*time^2

=0*1+0.5*5*1^2=2.5 m

hence height above ground=8+2.5=10.5 m

part c:

when the cables are cut, the elevator was moving upward at a speed of 5 m/s

after the cables are cut, its acceleration will be 9.8 m/s^2, downwards

so it will slow down while moving up, then stop momentarily when its speed becomes 0 and then speed up downwards

motion during slowing down before speed becomes 0:

initial velocity=5 m/s

acceleration=-9.8 m/s^2(as it is opposite to the direction of motion)

final velocity=0 m/s

time taken=(final velocity-initial velocity)/acceleration

=(0-5)/(-9.8)=0.5102 seconds...(1)

height risen during this time=5*0.5102-0.5*9.8*0.5102^2=1.2755 m

hence total maximum height above ground=10.5+1.2755=11.7755 m

motion during maximum height to ground:

initial speed=0

acceleration=9.8 m/s^2

distance=11.7755 m

then final speed=sqrt(initial speed^2+2*acceleration*distance)

=15.192 m/s

time taken=(final speed-initial speed)/acceleration=(15.192-0)/9.8=1.55 seconds..(2)

part d:

total time taken after the cables are cut=time taken to rise to maximum height+time taken to travel from maximum height to ground

=time obtained in equation 1+time obtained in equation 2
=0.5102+1.55=2.0602 seconds

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