Consider a large spring, hanging vertically, with spring constant k = 3220 N/m. If the spring is stretched 25.0 cm from equilibrium and a block is attached to the end, the block stays still, neither accelerating upward nor downward. What is the mass of the block?
answer should be in kg.
1
Expert's answer
2011-02-24T06:14:33-0500
The weight of the blok (mg) is balanced by the elactic force of spring (k Δx), thus mg = k Δx; m = k Δx/ g = 3220[N/m]* 0.25[m]/9.8[m/s2] = 82.14 [kg].
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