A weight of 0.2 kg is attached to the free end of a spring with a stiffness of 50 N / m located on a smooth table. The spring is stretched 0.25 m and then released. Find the highest speed of movement of the load (neglect the mass of the spring).
m = 0.2 kilograms is the mass of the load attached to the spring;
k = 50 N / m – coefficient of spring stiffness;
dx = 0.25 meters – the amount of tension of the spring.
It is required to determine v (m / s) – the maximum speed of the load.
Since, according to the condition of the problem, the load is located on a smooth table, we will not take into account the loss of energy due to the friction force.
The potential energy of a stretched spring is:
W = k * dx ^ 2/2 = 50 * 0.25 ^ 2/2 = 25 * 0.0625 = 1.6 Joules.
The body will have maximum speed when the potential energy of the stretched spring is completely converted into the kinetic energy of the body:
m * v ^ 2/2 = W;
m * v ^ 2 = 2 * W;
v ^ 2 = 2 * W / m;
v = (2 * W / m) ^ 0.5 = (2 * 1.6 / 0.2) ^ 0.5 = 16 ^ 0.5 = 4 m / s.
Answer: the maximum body speed is 4 m / s.
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