Answer:x=23.4 cm
Explanation:
Given
mass of block [tex]m=0.5 kg[/tex]
inclination [tex]\theta =30[/tex]
coefficient of static friction [tex]\mu =0.35[/tex]
coefficient of kinetic friction [tex]\mu _k=0.25[/tex]
distance traveled [tex]d=77.3 cm[/tex]
spring constant [tex]k=35 N/m [/tex]
work done by gravity+work done by friction=Energy stored in Spring
[tex]mg\sin \theta d-\mu _kmg\cos \theta d=\frac{kx^2}{2}[/tex]
[tex]mgd\left ( \sin \theta -\mu _k\cos \theta \right )=\frac{kx^2}{2}[/tex]
[tex]0.5\times 9.8\times 0.773\left ( \sin 30-0.25\cos 30\right )=\frac{35\times x^2}{2}[/tex]
[tex]x=\sqrt{\frac{2\times 0.5\times 9.8\times 0.773(\sin 30-0.25\times \cos 30)}{35}}[/tex]
[tex]x=0.234 m[/tex]
[tex]x=23.4 cm[/tex]
