Question

A 0.900 kg 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 48 cm/s.
Part A
What is the amplitude of the subsequent oscillations?
ANSWER:
Part B
What is the block's speed at the point where x = 0.25 A?

Answer 1

Answer:

Part A $x=0.1074m$

Part B $v=1.12m/s$

Explanation:

The force of the spring and the kinetic energy are the same it means that the amplitude of the subsequent give a velocity

$\frac{1}{2}*K*x^2=\frac{1}{2}*m*v^2$

Given: $m=0.9kg$, $K=18N/m$, $v=48cm/s*\frac{1m}{100cm}=0.48m/s$

Part A

The amplitude of the oscillations

$x^2=\frac{m*v^2}{K}$

$x=\sqrt{\frac{0.9kg*(0.48m/s)^2}{18N/m}}=\sqrt{0.01152m^2}$

$x=0.1074m$

Part B

The speed at the point x=0.25

$v^2=\frac{K*x^2}{m}$

$v=\sqrt{\frac{K*x^2}{m}}=\sqrt{\frac{18N/m*(0.25m)^2}{0.90kg}}=\sqrt{1.25m^2/s^2}$

$v=1.12m/s$