Question
Asked Nov 21, 2019
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A 1200 kg safe is 1.9 m above a heavy-duty spring when the rope holding the safe breaks. The safe hits the spring and compresses it 48 cm .

 

What is the spring constant of the spring?

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Expert Answer

Step 1

According to the conservation of energy, the loss of gravitational potential energy is equal to the gain in spring potential energy.

Rewrite the above relation for spring constant of the spring.

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mgh=kx2 2 2mgh k

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Step 2

Substitute the values, to fin...

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2(1200kg) (9.8m/s)(1.9m + 0.48 m) 0.01m (48 cm =242958.3N/m 103kN/m -243x10 N/m 1N/m 243 kN/m

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