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A horizontal spring is lying on a frictionless surface. Oneend
of the spring is attaches to a wall while the other end isconnected
to a movable object. The spring and object are compressedby 0.077
m, released from rest, and subsequently oscillate back andforth
with an angular frequency of 15.3 rad/s. What is the speed ofthe
object at the instant when the spring is stretched by0.033
m relative to its unstrained length?

Respuesta :

Answer:

[tex]v_{f}=1.0644 m/s[/tex]

Explanation:

According to law of conservation of energy:

Total initial Energy= Total Final Energy

[tex]E_{0} = E_{f}[/tex]

[tex]\frac{1}{2}*mv_{0} ^{2}+\frac{1}{2}*kx_{0} ^{2}=\frac{1}{2}*mv_{f} ^{2}+\frac{1}{2}*kx_{f} ^{2}[/tex]

In above expression  

[tex]v_{0}=0[/tex] (System is initially at rest)

So above expression will become:

[tex]\frac{1}{2}*kx_{0} ^{2}=\frac{1}{2}*mv_{f} ^{2}+\frac{1}{2}*kx_{f} ^{2}[/tex]

BY rearranging the above expression we will get final velocity/Speed:

[tex]v_{f}=\sqrt{\frac{k}{m}}*\sqrt{x_{0} ^{2}-x_{f} ^{2}}[/tex]

[tex]v_{f}=w*\sqrt{x_{0} ^{2}-x_{f} ^{2}}[/tex]

[tex]v_{f}=15.3*\sqrt{(-0.077)^{2}-(0.033)^{2}}[/tex]

[tex]v_{f}=1.0644 m/s[/tex]