contestada

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.

Respuesta :

berkayyakut96

Answer:

The answer is 41,4 [tex]\frac{m}{s^{2}}[/tex]

Explanation:

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

According to Newton's law [tex]F=ma[/tex], 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:

[tex]F_{spring} =kx[/tex]

Therefore [tex]F=360\frac{N}{m} .2,3.12^{-2} =8,28N[/tex]

And to find acceleration:

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

Otras preguntas