Amass of 2 kg is attached to a spring and placed on a horizontal surface. the spring has a spring constant of 20 n/m, and the spring is compressed 0.1 m past its natural length. if the mass is released from this compressed position, what is the speed of the mass as it passes the natural length of the spring?

The speed of the mass : v = 0.316 m/s

Further explanation

The energy used to press a spring is included as the potential energy

Can be formulated:

Ep= potential energy

k = spring constant

x = change in spring length

If the spring is released from its pressure, this potential energy will turn into kinetic energy

so applies the law of conservation of mechanical energy (Em)

Ek = Ep

A mass of 2 kg is attached to a spring, a spring constant of 20 N/m, and the spring is compressed 0.1 m past its natural length.

m = 2 kg

k = 20 N/m

x = 0.1 m

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Hooke's law

Keywords : spring,mass, spring constant,compressed position

Is a  vector quantity  that refers to "the rate at which an object changes its position." imagine a person moving rapidly - one step forward and one step back - always returning to the original starting position. while this might result in a frenzy of activity, it would result in a zero velocity. because the person always returns to the original position, the motion would never result in a change in position. since velocity is defined as the rate at which the position changes, this motion results in zero velocity. if a person in motion wishes to maximize their velocity, then that person must make every effort to maximize the amount that they are displaced from their original position. every step must go into moving that person further from where he or she started. for certain, the person should never change directions and begin to return to the starting position.