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It takes 87 j of work to stretch an ideal spring, initially 1. 4 m from equilibrium, to a position 2. 9 m from equilibrium. What is the value of the spring constant (force constant) of this spring?.

Respuesta :

LammettHash

The spring undergoes a change in potential energy of

1/2 k (2.9 m)² - 1/2 k (1.4 m)² = 87 J

Solve for k :

k ((2.9 m)² - (1.4 m)²) = 174 J

k = (174 J) / ((2.9 m)² - (1.4 m)²)

k ≈ 27 N/m

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