Question

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?

Answer 1

15.38 m.

Step-by-step 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 =
`V*\sqrt{(m)/(K) }`

Putting the values

x =
`79.856*\sqrt{(.717)/(19.32) }`

= 15.38 m.