A weight of 0.2 kg is suspended on a spring with a stiffness of 98 N / m. How many millimeters will the spring lengthen

A weight of 0.2 kg is suspended on a spring with a stiffness of 98 N / m. How many millimeters will the spring lengthen if the load begins to move along a circle with a radius of 0.6 m in the horizontal plane with a constant velocity of 2.1 m / s?

m = 0.2 kg.

k = 98 N / m.

R = 0.6 m.

V = 2.1 m / s.

x -?

Let us write Newton’s 2 law for the rotation of a body in the horizontal plane: m * a = Fcont, where m is the mass of the body, a is the centripetal acceleration, Fcont is the force that acts on the load.

Since the weight is attached to the spring, the elastic force from the side of the spring acts on the weight.

We find the elastic forces according to Hooke’s law: F = k * x.

We express the centripetal acceleration by the formula: a = V2 / R.

m * V ^ 2 / R = k * x.

x = m * V ^ 2 / R * k.

x = 0.2 kg * (2.1 m / s) ^ 2 / 0.6 m * 98 N / m = 0.015 m.

Answer: when rotating, the spring will lengthen by x = 0.015 m.