Answer:
240.1 N/m
Explanation:
Applying the formula for the potential energy stored in a stretched spring,
E = 1/2ke².................... Equation 1
Where E = potential energy, k = spring constant, e = extension.
make k the subject of the equation,
k = 2E/e².................. Equation 2
Given: E = 35 J, e = 0.54 m
Substitute into equation 2
k = 2(35)/0.54²
k = 70/0.2916
k = 240.1 N/m.
Hence the spring constant of the spring is 240.1 N/m
