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The 2-kg collar is attached to a spring that has an un-stretched length of 3.0 m. If the collar is drawn to point B and releases from rest, what is the speed when it arrives at point A. Note that k = 3.0 N/m and neglect friction.

Answer :

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Answer:

[tex]v = 3.4 \ m/s[/tex]

Explanation:

From the question we are told that

   The mass of the collar is  [tex]m = 2 \ kg[/tex]

    The original length is  [tex]L = 3.0 \ m[/tex]

     The spring constant is  [tex]k = 3.0 \ N/m[/tex]

     

Generally the extension of the spring  is  mathematically evaluated as

        [tex]e = 4 -3 = 1 \ m[/tex]

Now with Pythagoras theorem we can obtain the length from A to B as

        [tex]AB = \sqrt{5 ^2 + 4^2}[/tex]

       [tex]AB = 6.4 \ m[/tex]

The  extension of the spring at B is  

     [tex]e_b = 6.4 - 3 = 3.4 \ m[/tex]

According to the law of energy conservation

   The energy stored in the spring at point A +  the kinetic energy of the  spring =  The  energy stored on the spring at B

So

     [tex]\frac{1}{2} * k * e + \frac{1}{2} * m* v^2 = \frac{1}{2} * k * e_b[/tex]

substituting values

    [tex]\frac{1}{2} * 3 * 1^2 + \frac{1}{2} * 2* v^2 = \frac{1}{2} * 3 * 3.4^2[/tex]

=>   [tex]v = 3.4 \ m/s[/tex]

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