Physics
posted by Janelle on .
A mass of 6.10 kg rests on a smooth surface inclined 39.0o above the horizontal. It is kept from sliding down the plane by a spring attached to the wall. the spring is aligned with the plane and has a spring constant of 126 N/m. How much does the spring stretch?

can you show the steps to how to answer the question as well ? just the steps would be fine :)

vector m•a=vector m•g +vector F(spring)
a=0 =>
0=vector m•g +vector F(spring)
x: 0 = m•g•sinα –k•x,
x=m•g•sinα/k=6.1•9.8•sin 39°/126 =0.3 m 
Fm = M*g = 6.10 * 9.8 = 59.8 N. = Force of the mass.
Fp = 59.8*sin39 = 37.63 N. = Force parallel to the incline = Force applied to the spring.
d=37.63/126 * 1m = 0.30 m. = Distance the spring is stretched.