The cable of 1500 kg elevator in figure below snaps when the elevator is at rest

Question

The cable of 1500 kg elevator in figure below snaps when the elevator is at rest at first floor so that the bottom of the elevator is a distance 4m above cushioning spring whose spring constant is k=1.5 times 10^5 N/m. A safety device clamps the guide rails so that a constant friction force of 4000N opposes the motion of the elevator. Find the speed of the elevator just before it hits the spring Find the distance that the spring is compressed. Find the distance that the elevator will bounce back up above uncompressed spring. use the conservation of energy principle to find the coming to rest. (Consider both cases that it comes to rest while upward or was moving downward.)1/2(1500)v^2-1500 times 9.8 times 4=-(4000) times 4

Explanation / Answer

a)speed of the elevator v=?

(1/2)mv2-mgh=-Fh here m=1500 kg, h=4m,g=9.8m/s2,F=4000 N

1/2 1500*v2-1500*9.8*4=-4000*4

v=7.55 m/s

b)compressing distance x=?

1/2 kx2-1/2 mv2-mgx=-Fh

1/2 1.5*105x2-1/2 1500*7.552=1500*9.8*x=-4000*x

x=0.83 m

c)bouncing distance y=?

1/2kx2=mgy

1/2*1.5*105*(0.83)2=1500*9.8*y

y=13.18 m

d)t0tal distance =compressing distance+bouncing distance

=0.83+13.18=14.01 m

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