Answer:
0.1117 m
Explanation:
Given information
m=12 kg, d=3m , [tex]\theta=32^{\circ}[/tex], [tex]k =3.0*10^{4} N/m[/tex]
Net work done W = Fd = mgdcos(90-32) and taking g as 9.81
W = 12*9.81*3*sin(32) = 187.1463 J
From the principal of conservation of energy
W = change in potential energy of spring
[tex]W = 0.5kx^{2}[/tex]
[tex]187.1463 = 0.5(3.0*10^{4})x^{2}[/tex]
[tex]x=\sqrt {\frac {187.1463}{0.5*(3*10^{4})}[/tex]
[tex]x=\sqrt {0.012476}=0.111698\approx 0.1117 m[/tex]
