In the worst-case scenario, you've got an elevator with a mass 2420 kg moving a speed...

In the worst-case scenario, you've got an elevator with a mass 2420 kg moving a speed 4.29 m/s when it reaches the bottom of the shaft. Your first idea is to put a spring, with spring constant 1.068×10⁴ N/m, at the bottom of the shaft.

a)Assuming the spring obeys Hooke's Law perfectly, how far will it compress to stop the elevator?

b)When the elevator comes to a stop with the spring compressed, what is the net force on the elevator?

c)It's a bit unreasonable to put a spring large enough to compresses this far under every elevator. So you decide to add a clamping mechanism to the walls around the spring, which will provide a friction force of magnitude 1.74×10⁴ N

Now how far does the spring get compressed?

Express your answer with appropriate units.

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Expert Solution

genius_generous answered 1 week ago

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