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Suppose a mass was stretching a vertical spring. Why does the mass have a velocity in its equilibrium position where the net force is 0?

Note the spring is massless.

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4 Answers 4

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Net force is zero does not mean velocity is zero! It simply means acceleration is zero.

At its equilibrium, the elastic force of the spring ($kx$->Hooke's Law) is equal to the weight ($mg$) of the object, for a vertically oscillating spring.

In a simple harmonic motion, energy is conserved, and is equivalent to the sums of the kinetic energy, gravitational potential energy (for vertical spring) and the elastic potential energy. As the box moves away from the equilibrium, kinetic energy is converted to other forms of energy. In fact, it is at the equilibrium where kinetic energy is the highest.

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Suppose you released the spring mass system from some initial compression x . As the spring is compressed it tries to push the block . In this process the spring looses it's potential energy and the block suffers a loss in potential energy . And according to law of conservation of energy and law of conservation of mechanical energy this has to go into blocks kinetic energy . So even when the block is at its mean position it still has some kinetic energy to satisfy the law of conservation of energy.

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The net force on the block is zero which simply implies its acceleration at mean position is zero. However, the |v| remains the same and it is maximum at mean position. It can also be noted from from v vs t graph , where the velocity is maximum and constant implies, its time derivative will be zero ($$ a= {dv}/{dt}=0 $$) and hence the net force is also zero.

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From Newton's second law $F=ma$, the net force equal to zero implies that the acceleration (rate of change of velocity) is zero but not that the velocity is zero.
If you have studied simple harmonic motion you will know that when a body passes through the static equilibrium position the net force and hence the acceleration of the body is zero and yet its kinetic energy and hence its speed is a maximum.

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  • $\begingroup$ So the speed is at a maximum when the block is in an equilibrium position and the derivative of the v vs t graph is zero? $\endgroup$
    – Keeper
    Commented Nov 19, 2018 at 4:51

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