Two identical carts weighing 100 g each are connected by a compressed spring. The length of the spring is 6 cm.

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

m1 = 100 g = 0.1 kg.

l1 = 6 cm = 0.06 m.

k = 30 N / m.

a1 = 6 m / s2.

l0 -?

Since the trolleys have the same mass, when the spring interacts with the trolleys, the elastic force of the spring Fпр imparts the same acceleration to the carts: Fпр = 2 * m1 * а1.

We will express the force of elasticity of the spring Fcontrol by Hooke’s law: Fcont = k * (l0 – l1), where l0 is the length of the undeformed spring, l1 is the length of the deformed spring.

k * (l0 – l1) = 2 * m1 * a1.

l0 – l1 = 2 * m1 * a1 / k.

l0 = l1 + 2 * m1 * a1 / k.

l0 = 0.06 m + 2 * 0.1 kg * 6 m / s2 / 30 N / m = 0.1 m.

Answer: the undeformed spring had a length l0 = 0.1 m.