Answer:
Compression of the spring: 0.18 m (downward)
Explanation:
The forces acting on the block of wood are:
- The force of gravity, acting downward, of magnitude [tex]mg[/tex], where m = 5.32 kg is the mass of the block and [tex]g=9.8 m/s^2[/tex] is the acceleration due to gravity
- The force exerted by the spring, downward, of magnitude [tex]kx[/tex], where [tex]k=179N/m[/tex] is the spring constant and [tex]x[/tex] is the elongation of the spring
- The buoyant force, upward, of magnitude [tex]\rho V g[/tex], where [tex]\rho=1000 kg/m^3[/tex] is the water density and V the volume of the block
Since the block is in equilibrium, the net force is zero, so we can write
[tex]mg+kx-\rho V g=0[/tex] (1)
We have to find the volume of the block first. We have:
m = 5.32 kg (mass)
[tex]\rho_w = 622 kg/m^3[/tex] (wood density)
So, the volume is
[tex]V=\frac{m}{\rho_w}=\frac{5.32}{622}=0.0086 m^3[/tex]
So now we can re-arrange eq.(1) to find the elongation of the spring, x:
[tex]x=\frac{-mg+\rho Vg}{k}=\frac{-(5.32)(9.8)+(1000)(0.0086)(9.8)}{179}=0.18 m[/tex]
So, the spring is compressed by 0.18 m.
