A 1.10kg block is attached to a spring with spring constant 18 N/m. While the block is sitting at rest, a student hits it with a hammer and almost instantaneously gives it a speed of 40cm/s .What is the block's speed at the point where x = 0.45 A?

Respuesta :

Answer:

The velocity is  [tex]v_x = 0.356 \ m/s[/tex]

Explanation:

From the question we are told that

   The  mass is  [tex]m = 1.10 \ kg[/tex]

   The  spring constant is  [tex]k = 18 \ N/m[/tex]

   The  speed is [tex]v = 40 \ cm / s = 0.4 m/s[/tex]

    The  position considered is  x =  0.45 A  

Here A is the amplitude which is mathematically represented as

     [tex]A = v * \sqrt{\frac{m}{k} }[/tex]

=>     [tex]A = 0.4 * \sqrt{\frac{1.10}{18 } }[/tex]

=>     [tex]A = 0.0989 \ m[/tex]

So     [tex]x = 0.45 * 0.0989[/tex]

=>     [tex]x = 0.045 \ m[/tex]

Generally the speed at  x  is mathematically represented as

      [tex]v_x = \sqrt{ \frac{k}{m} * [A^2 - x^2 ]}[/tex]

=>    [tex]v_x = \sqrt{ \frac{18}{ 1.10} * [0.0989^2 - 0.045^2 ]}[/tex]

=>    [tex]v_x = 0.356 \ m/s[/tex]