Two identical trolleys, each weighing 100 g, are interconnected by a compressed spring.

Two identical trolleys, each weighing 100 g, are interconnected by a compressed spring. The length of the spring (compressed) is 6 cm. The spring rate is 30 N / m. After the spring was unclenched, the carts parted with an acceleration of 6 m / s2. Find the length of the undeformed spring.

m1 = m2 = 100 g = 0.1 kg.

l = 6 cm = 0.06 m.

a1 = a2 = 6 m / s2.

k = 30 N / m.

l0 -?

The force of elasticity of the spring Fпр acts on the carts and imparts to them the accelerations a1 and a2.

Fupr = m1 * a1 + m2 * a2.

We express the elastic force according to Hooke’s law: Fel = k * (l – l0), where k is the stiffness of the spring, l – l0 is the change in its length.

k * (l0 – l) = m1 * a1 + m2 * a2.

k * l0 – k * l = m1 * a1 + m2 * a2.

l0 = (k * l + m1 * a1 + m2 * a2) / k.

l0 = (30 N / m * 0.06 m + 0.1 kg * 6 m / s2 + 0.1 kg * 6 m / s2) / 30 N / m = 0.1 m.

Answer: the initial length of the spring is l0 = 0.1 m.