A 0.16kg block on a horizontal frictionless surface is attatched to a spring who

Question

A 0.16kg block on a horizontal frictionless surface is attatched to a spring whose force constant is 360 N/m. The block is pulled from tis equilibrium position at x=0.0 m to a dispacement x = +0.080 m and is released from rest. The block then exectues simple harmonic motion along the x-axis (horizontal). When the displacement is x=-0.045 m, the acceleratoin of the block is closest to?

The answer is 100 m/s^2. I don't know how to get this answer tho so I need someone to show the work needed to get it.

Thanks.

Explanation / Answer

F = Ma = Kx = 360* 0.08m =28.8 N => a = 28.8/0.16 = 180m/s^2.

Initial acceleration = 180m/s^2.

We know in SHM, a(x) = -w^2x,

so here, a1/a2 = x1/x2 => a2 = (0.045/0.080)*180 = 101.25 m/s^2.

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