A spring with spring constant of 26 N/m is stretched 0.22 m from its equilibrium position. How much work must be done to stretch it an additional 0.12 m

1.503 J

Explanation:

Work done in stretching a spring = 1/2ke²

W = 1/2ke² Equation 1

Where W = work done, k = spring constant, e = extension.

Given: k = 26 N/m, e = (0.22+0.12), = 0.34 m.

Substitute into equation 1

W = 1/2(26)(0.34²)

W = 13(0.1156)

W = 1.503 J.

Hence the work done to stretch it an additional 0.12 m = 1.503 J

true tan speed = ang vel x radius

i think its c if i'm wrong