# Answer on Mechanics | Relativity Question for reyne

Question #1656

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.

answer should be in kg.

Expert's answer

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/s

mg = k Δx;

m = k Δx/ g = 3220[N/m]* 0.25[m]/9.8[m/s

^{2}] = 82.14 [kg].Need a fast expert's response?

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