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A 0.37 kg object is attached to a spring with a spring constant 175 N/m so that the object is allowed to move on a horizontal frictionless surface. The object is released from rest when the spring is compressed 0.19 m. Find the force on the object when it is released. Answer in units of N. What is the acceleration at this instant? Answer in units of m/s^2.​

Respuesta :

Answer:

[tex]F=33.25\ N[/tex]

[tex]a=89.8649\ m.s^{-2}[/tex]

Explanation:

Given:

  • mass of the object, [tex]m=0.37\ kg[/tex]
  • spring constant, [tex]k =175\ N.m^{-1}[/tex]
  • compression in the spring, [tex]\Delta x=0.19\ m[/tex]

Now the force on the spring on releasing the compression:

[tex]F=k. \Delta x[/tex]

[tex]F=175\times 0.19[/tex]

[tex]F=33.25\ N[/tex]

Now the acceleration due to this force:

[tex]a=\frac{F}{m}[/tex]

[tex]a=\frac{33.25}{0.37}[/tex]

[tex]a=89.8649\ m.s^{-2}[/tex]

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