A 3.0-kilogram mass is attached to a spring having a spring constant of 300 N/m. The spring is oriented horizontally. The mass is pulled 0.20 meters from the spring's equilibrium position and released from rest. Ignoring friction, what is the maximum velocity achieved by the spring-mass system (in m/s)?

When the spring is released, the body has a maximum velocity of 5.477 m/s.

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The potential energy of a spring is given by the relation PEs = (1/2)*k*x^2, where k is the spring constant, and x is the displacement of the spring from its equilibrium position.

In the problem, the spring constant is equal to 300 N/m and the displacement is equal to 0.2...

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The potential energy of a spring is given by the relation PEs = (1/2)*k*x^2, where k is the spring constant, and x is the displacement of the spring from its equilibrium position.

In the problem, the spring constant is equal to 300 N/m and the displacement is equal to 0.2 m.

Using the relation provided earlier, the potential energy of the spring is (1/2)*300*0.2 = 30 J.

A 3.0 kilogram mass is attached to the spring. If friction can be ignored, the kinetic energy of the mass is equal to the potential energy of the spring at the point where its velocity is the highest. The kinetic energy of a body of mass m traveling at a velocity v is given by (1/2)*m*v^2. Here, m is equal to 3.0 and v has to be determined.

Equating the two gives: (1/2)*k*x^2 = (1/2)*m*v^2

=> 30 = (1/2)*m*v^2

=> 30 = (1/2)*2*v^2

=> 30 = v^2

=> v = sqrt 30

=> v = 5.477 m/s

The maximum velocity of the 3 kg mass is 5.477 m/s.

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