A 0.28-kg block on a horizontal frictionless surface is attached to an ideal massless spring whose spring constant is 500 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 = -0.052 m, find the acceleration of the block.

The block will accelerate at 92.86m/s²

Explanation:

The acceleration of a simple harmonic motion of a spring is expressed as

a= - kx/m

Where k = spring constant

x= displacement

m= mass of block

Given data

Spring constant k = 500N/m

Displacement x= - 0.052m

Mass of block m= 0.28kg

Pluging this parameters into the expression for acceleration we have

a= - 500*(-0.052)/0.28

a= 26/0.28

a= 92.86m/s²

you get 2.3=1.21*\frac{a}2 or 4.6=1.21a or a=3.802 (b)

explanation: