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A 0.16-kg block on a horizontal frictionless surface is attached to an ideal spring whose force constant (spring constant) is 360 N/m. The block is pulled from its equilibrium position at x = 0.000 m to a position x = +0.080 m and is released from rest. The block then executes simple harmonic motion along the horizontal x-axis. When the position is x = -0.037 m, what is the acceleration of the block? 8

Respuesta :

Answer:

[tex]a = 83.25 m/s^2[/tex]

Explanation:

As we know that the mean position of the block is given as x = 0

so when block is at x = -0.037 m then compression in the spring is given as

[tex]\Delta x = (x_i - x_f)[/tex]

[tex]\Delta x = 0 + 0.037 = 0.037 m[/tex]

now we know that spring force is given as

[tex]F = k\Delta x[/tex]

[tex]F = 360(0.037)[/tex]

[tex]F = 13.32 N[/tex]

Now the acceleration of the block is given as

[tex]a = \frac{F}{m}[/tex]

[tex]a = \frac{13.32}{0.16}[/tex]

[tex]a = 83.25 m/s^2[/tex]

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