A0.20-kg block on a horizontal frictionless surface is attached to an ideal massless spring whose spring constant is 360 n/m. the block is pulled from its equilibrium position at x = 0.00 m to a displacement x = +0.080 m and is released from rest. the block then executes simple harmonic motion along the horizontal x-axis. when the displacement is x = -2.3×10−2 m, find the acceleration of the block.

The answer is 41,4

Explanation:

To understand this problem we need to visualize when block is at the equilibrium position and when the displacement is x=.

According to Newton's law , if we find the force applied to the block, we can find the acceleration since we know the mass of the block.

If you compress a spring by x displacement under F force, the spring's force can be found as:

Therefore

And to find acceleration:

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6 cos 34, 6 sin 34 trig