The cable of a 4200 lb elevator in the figure below snaps when the elevator is a

Question

The cable of a 4200 lb elevator in the figure below snaps when the elevator is at rest at the first floor so that the bottom is a distance d = 12.0 ft above a cushioning spring whose force constant is k = 10,000 lb/ft. A safety device clamps the guide rails, removing 1000 ft-lb of mechanical energy for each 1.00 ft that the elevator moves.

(a) Find the speed of the elevator just before it hits the spring in ft/s.

(b) Find the distance that the spring is compressed.  

Step by step explanation would be helpful.

Explanation / Answer

From the given question,

mass= 4200lb

distance travelled=12 ft

force constant ( k) = 10,000 lb/ft

Total potential energy= 4200*12 ft lb=50400 ft lb

energy removed=12000 ft lb

energy remaining=50400-12000=38400 ft lb

kinetic energy= (1/2)(4200/32)v2=38400

solving we get v=29.2 ft/s

speed of the elevator just before it hits the spring is 29.2 ft/s

Let x be the distance spring is compressed.

38400=(1/2)kx2

38400=(1/2)(10000)x2

x=2.77 ft

The spring is compressed by 2.77 ft

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