A block slides with no friction and hits a spring with spring constant k=2000 N/m. The block compresses the spring in a straight line for a distance 0.15m. The block’s kinetic energy, in J, at that point is 0 J. What was its initial kinetic energy, in Joules?

Respuesta :

To solve this exercise it is necessary to apply the concepts related to kinetic energy in a spring.

By definition the kinetic energy keeping what is stipulated in the stable Hooke law that

[tex]KE_s = \frac{1}{2} kx^2[/tex]

Where,

K = Spring constant

x = Displacement

Our values are given as,

x = 0.15m

k = 2000N/m

Therefore replacing

[tex]KE_s = \frac{1}{2} kx^2[/tex]

[tex]KE_s = \frac{1}{2} (2000)(0.15)^2[/tex]

[tex]KE_s = 22.5J[/tex]

Therefore the initial kinetic energy is 22.5J

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