You exert a 100-N pull on the end of a spring. When you increase the force by 20% to 120 N, the spring's length increases 4.0 cm beyond its original stretched position. Part A What is the spring constant of the spring

Spring constant of the spring = 500 N/m

Explanation:

Hooke's law for a spring explains that, provided the elastic limit isn't exceeded,

F = kx

where = Force or Load on a spring in Newton

k = spring constant

x = extension or compression of the spring

When the force is 100 N, let the extension be x m

100 = kx

x = (100/k) eqn 1

When the force is 120 N, the extension becomes (x + 4 cm) = (x + 0.04) m

120 = k(x + 0.04)

120 = kx + 0.04k eqn 2

Substitute the value of x obtained in eqn 1 into eqn 2

120 = k(100/k) + 0.04k

120 = 100 + 0.04k

0.04k = 20

k = (20/0.04)

k = 500 N/m

Hope this Helps!!!