Explanation:
The given data is as follows.
Spring constant (k) = 78 N/m, [tex]\theta = 30^{o}[/tex]
Mass of block (m) = 0.50 kg
According to the formula of energy conservation,
mgh sin [tex]\theta = \frac{1}{2}kx^{2}[/tex]
h = [tex]\frac{1}{2} \times \frac{kx^{2}}{mg Sin \theta}[/tex]
= [tex]\frac{78 \times 0.04}{2 \times 0.5 \times 9.8 \times 0.5}[/tex]
= 0.64 m
Thus, we can conclude that the distance traveled by the block is 0.64 m.
