Question

A 0.20-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.

Answer 1

Answer:

The answer is 41,4 $\frac{m}{s^{2}}$

Explanation:

To understand this problem we need to visualize when block is at the equilibrium position and when the displacement is x=$-2,3.10^{-2}$.

According to Newton's law $F=ma$, 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:

$F_{spring} =kx$

Therefore $F=360\frac{N}{m} .2,3.12^{-2} =8,28N$

And to find acceleration:

$F=ma$ → $8,28=0,20a$ → $a=41,4\frac{m}{s^{2}}$