Four bars weighing 100 grams are suspended on a spring of 15 cm tension. Determine the spring tension by adding a similar bar.

Let’s find the total mass of the cargo:

0.1 * 4 = 0.4 kg.

Then the force of gravity will be:

F = m * g = 0.4 * 10 = 4 n.

The coefficient of spring stiffness will be:

k = F / Δl = 4 / 0.15 = 400/15 = 27.

After attaching another bar, the mass will become equal to:

0.4 + 0.1 = 0.5.

Then the tension of the spring will be equal to:

Δl = F / k = 10 * 0.5 / 27 = 0.185 m.

Answer: The stretch amount will be 0.185 m.