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A 0.40 kg object is attached to a spring with force constant 157 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 0.17 m.
(a) Find the force on the object.
(b) Find its acceleration at that instant.

Respuesta :

Answer:

Part a)

[tex]F = 26.7 N[/tex]

Part b)

[tex]a = 66.7 m/s^2[/tex]

Explanation:

Part a)

Force on the object due to spring force is given as

[tex]F = kx[/tex]

here we know that

[tex]k = 157 N/m[/tex]

[tex]x = 0.17 m[/tex]

so we have

[tex]F = 157\times 0.17[/tex]

[tex]F = 26.7 N[/tex]

Part b)

Acceleration of the object is given as

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

[tex]a = \frac{26.7}{0.40}[/tex]

[tex]a = 66.7 m/s^2[/tex]

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