A dynamometer and a ruler are at your disposal. Stretching the dynamometer spring by 5 cm

A dynamometer and a ruler are at your disposal. Stretching the dynamometer spring by 5 cm, you find that its reading is equal to 2H. What is the stiffness of the dynamometer spring?

Given:

dx = 5 centimeters – the amount by which the dynamometer spring was stretched;

F = 2 Newton – dynamometer readings after stretching the spring.

It is required to determine k (Newton / kilogram) – the coefficient of stiffness of the dynamometer spring.

We translate the units of measurement of length into the SI system:

dx = 5 centimeters = 5 * 10-2 = 5/100 = 0.05 meters.

Then, to determine the stiffness coefficient, you must use the following formula:

F = k * dx, from here we find that:

k = F / dx = 2 / 0.05 = 200/5 = 40 Newton / meter.

Answer: the coefficient of stiffness of the dynamometer spring is 40 Newton / meter.