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Consider a block attached to horizontal spring that moves along a horizontal, frictionless surface. The system is the block and the spring. If the block is released from rest when the spring is stretched by a displacement of 4.50cm from its unstrained length, determine the displacement of the spring from its unstrained length when its translational speed is 2.00m/s. The mass of the block is 0.200kg and the spring constant of the spring is k=545N/m.

Respuesta :

Answer:

x = 0.023m

and:

x = -0.023m

Explanation:

Using the conservation of energy:

[tex]E_i = E_f[/tex]

so:

[tex]\frac{1}{2}KX_i^2=\frac{1}{2}KX_f^2+\frac{1}{2}MV^2[/tex]

where K is the constant of the spring, Xi is the initial displacement, Xf the displacement when the speed is 2m/s, M the mass of the block and V the velocity of the system.

So, replacing values, we get:

[tex]\frac{1}{2}(545)(0.045m)^2=\frac{1}{2}(545)X^2+\frac{1}{2}(0.2kg)(2m/s)^2[/tex]

solving for Xf, we get:

Xf = 0.023m

and:

Xf = -0.023m

So, the displacement of the spring from its unstrained length when the speed is 2m/s is 0.023m

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