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/s2] = 82.14 [kg].
mg = k Δx;
m = k Δx/ g = 3220[N/m]* 0.25[m]/9.8[m/s2] = 82.14 [kg].
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