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a heavy block is suspended from a vertical spring. the elestic potential energy is stored in the spring is 2 j. what is the spring constant if the elongation of the spring is 10 cm

Respuesta :

Answer:

k = 100 N/m

Explanation:

The elastic potential energy of a spring is equal to (1/2)kx^2, where k is the spring constant and x is the distance compressed. If the elastic PE is equal to 2J, we can solve for k. Make sure to convert units from cm to m:

(1/2)kx^2 = 2J

kx^2 = 1

k = 1/x^2

k = 1/(0.1)^2

k = 100 N/m

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