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?

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

the force on the thing and the distance the force moves the thing

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