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A 0.717kg block is attached to a spring with spring constant 19.32N/m. While the block is sitting at rest, a student hits it with a hammer and almost instantaneously gives it a speed of 79.856cm/s. What is the amplitude of the subsequent oscillations in meters?

Respuesta :

Answer:

15.38 m.

Explanation:

The kinetic energy of the block is equal to potential energy of spring at maximum compression

1/ 2 m V² = 1 /2 K X²

m is mass of block , V is its velocity , K is spring constant and  X is maximum compression or its amplitude.

X = [tex]V\times\sqrt{\frac{m}{K} }[/tex]

Putting the values

x = [tex]79.856\times\sqrt{\frac{.717}{19.32} }[/tex]

= 15.38 m.