A 400g weight vibrates on a spring with a stiffness of 250N / m. The vibration amplitude is 15cm.

| March 4, 2021 | education

A 400g weight vibrates on a spring with a stiffness of 250N / m. The vibration amplitude is 15cm. Find the total mechanical vibration energy and the maximum speed. In which position is it reached?

m = 400 g = 0.4 kg.
k = 250 N / m.
x = 15 cm = 0.15 m.
E -?
V -?
The total mechanical energy E is the sum of the potential energy En and the kinetic energy Ek.
E = En + Ek.
The potential energy of a spring pendulum is determined by the formula: En = (k * x ^ 2) / 2.
The kinetic energy of a spring pendulum is determined by the formula: Ek = (m * V ^ 2) / 2.
When the body is in the extreme position, it stops. Then the total energy is added only from potential energy: E = En.
E = (k * x ^ 2) / 2.
E = (250 N / m * (0.15 m) ^ 2) / 2 = 2.81 J.
When the load passes the equilibrium position, then its amplitude is x = 0. This means that in the equilibrium position, the total energy will be added only from kinetic energy. At this moment, the speed will be maximum.
E = Ek.
E = (m * V ^ 2) / 2.
V ^ 2 = 2 * E / m.
V = √ (2 * E / m).
V = √ (2 * 2.81 J / 0.4 kg) = 3.75 m / s.
Answer: total energy E = 2.81 J, maximum speed of the load V = 3.75 m / s at the moment of passing the equilibrium position.



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