How much force is required to pull a spring 3.0 cm from its equilibrium position if the spring constant is 3.7 x 103 n/m?

Hooke's law states that for a helical spring the extension is directly proportional to the force applied provided the elastic limit is not exceeded.

Therefore; F = ke, where k is the spring constant, F is the force and e is the extension.

k = 2700 N/m and e = 3 cm or 0.03 M

therefore, F = 2700 * 0.03

= 81 N

Thus, the force required will be 81 N

From Hooke's Law

We have F = ke where k is the spring constant and e is the extension.

So e = 3cm = 0.03m and k is 3.7 * 10^3

So it follows that F = 3.7 * 10^3 * 0.03 = 3.7 * 10^3 * 3 * 10^-2

So F = 11.1 * 10^ (3-2) = 11.1 * 10 = 111N