If a spring has an elastic potential energy of 60 J and a displacement of 4 m , what is the spring constant of the spring?
1 answer:
Answer:
7.5 N/m
Explanation:
Potential energy of a spring can be calculated using below formula
Potential energy= 1/2kx^2
potential energy = 60 J
X= displacement = 4 m
K= spring constant=?
Substitute the values we have
60= 1/2 × K × 4^2
60= 1/2 × K × 16
60= K × 8
K= 7.5 N/m
Hence, the spring constant of the spring is 7.5 N/m
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