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A 3.7kg block on a horizontal frictionless surface is attached to a spring whose force constant is 450 N/m. The block is pulled from equilibrium position at x=0m to a displacement x=+0.080m and is released from rest. The block then executes simple harmonic motion along the x-axis. The maximum elastic potential energy of the system is closest to? (Answer: 1.4 J) I just need to understand how to get there. Please show the equations used

Respuesta :

Answer:

U=1.44 J

Explanation:

Given that

m = 3.7 kg

K = 450 N/m

X= 0.08 m

We know that spring stored potential energy(U) and this potential energy given as

[tex]U=\dfrac{1}{2}KX^2[/tex]

By putting the values

[tex]U=\dfrac{1}{2}KX^2[/tex]

[tex]U=\dfrac{1}{2}\times 450\times 0.08^2[/tex]

U=1.44 J

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