Physics 8.7 fe question

Question

A spring exerts a 16 N force after being stretched by 1.0 cm from its equilibrium length.

By how much will the spring force increase if the spring is stretched from 5.0 cm away from equilibrium to 7.0 cm from equilibrium? (in N)

Answer 1

If k is the spring constant and T the tension:

T = kx

16 = k * 1

k = 16 N/cm.

When x = 5 cm, T1 = 16 * 5 N

When x = 7 cm, T2 = 16 * 7 N.

The increase is T2 - T1

= 32 N.