The block will oscillate back and forth along the horizontal surface with a period given by the equation $T = 2\pi\sqrt{\frac{m}{k}}$,
where $m$ is the mass of the block, $k$ is the force constant of the spring, and $T$ is the period of the oscillation. In this case,
the mass of the block is 150 g, which is equivalent to 0.15 kg, and the force constant of the spring is 3.2 N/m.
Plugging these values into the equation, we get
$T = 2\pi\sqrt{\frac{0.15}{3.2}} \approx 0.3268\text{ s}$
The amplitude of the oscillation is equal to the maximum displacement of the block from its equilibrium position, which in this case is 3.0 cm. The block will oscillate back and forth with this amplitude and with a period of approximately 0.3268 seconds.
Learn more about equilibrium:
https://brainly.com/question/28467329
#SPJ4
