An aluminum bar measuring 2 * 4 * 10 cm is suspended on a spring with a stiffness of 200 N / m.

An aluminum bar measuring 2 * 4 * 10 cm is suspended on a spring with a stiffness of 200 N / m. How much is the spring extended?

Given:

k = 200 N / m (Newton per meter) – the stiffness of the spring to which the aluminum bar is suspended;

a = 2 centimeters = 0.02 meters – the height of the bar;

b = 4 centimeters = 0.04 meters – the width of the bar;

c = 10 centimeters = 0.1 meter – the length of the bar;

ro = 2700 kg / m3 (kilogram per cubic meter) – the density of aluminum;

g = 10 Newtons per meter – acceleration due to gravity (rounded off value).

It is required to determine dx (meter) – spring elongation.

Let’s find the volume of the bar:

V = a * b * c = 0.02 * 0.04 * 0.1 = 0.00008 m3.

The mass of the bar will be equal to:

m = ro * V = 0.00008 * 2700 = 0.216 kilograms.

The force of gravity acting on the bar will be equal to:

F = m * g = 0.216 * 10 = 2.16 Newton.

Then the elongation of the spring will be equal to:

dx = F / k = 2.16 / 200 = 0.001 meter = 1 millimeter.

Answer: the spring will lengthen by 1 millimeter.