Two trolleys weighing 100 and 150 g are connected by an associated spring.

Two trolleys weighing 100 and 150 g are connected by an associated spring. When the spring was released, the acceleration of the first cart was 4.5 m / s ^ 2. With what force does the spring act on the second cart?

m1 = 100 g = 0.1 kg.

m2 = 150 g = 0.15 kg.

a1 = 4.5 m / s2.

F2 -?

a2 -?

The spring, when deflated, acts on bogies of mass m1 and m2 with the same force F: F1 = F2.

According to 2 Newton’s law, the force F, which acts on a body, is equal to the product of the body’s mass m by the body’s acceleration a: F = m * a.

F1 = m1 * a1.

F2 = m1 * a1.

F2 = 0.1 kg * 4.5 m / s2 = 0.45 N.

m1 * a1 = m2 * a2.

a2 = m1 * a1 / m2.

a2 = 0.1 kg * 4.5 m / s2 / 0.15 kg = 3 m / s2.

Answer: force F2 = 0.45 N acts on the second body, it will move with acceleration a2 = 3 m / s2.