Question

A mass 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?

Answer 1

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:

$\displaystyle E_p=\frac{1}{2}kx^2$

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

$\displaystyle Ep=Ek\\\\\frac{1}{2}kx^2=\frac{1}{2}mv^2\\\\kx^2=mv^2\\\\20\times0.1^2=2\times v^2\\\\v^2=0.1\\\\v=\sqrt{0.1}\\\\v=0.316~m/s$

Learn more

Hooke's law

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Keywords : spring,mass, spring constant,compressed position