Answer:
the elongation of the spring is 0.3186 m.
Step-by-step explanation:
The force exerted by the spring on the block of wood when it is in equilibrium is equal to the weight of the block of wood, which is given by:
F_spring = F_gravity
where F_spring is the force exerted by the spring and F_gravity is the weight of the block.
The weight of the block of wood can be calculated as:
F_gravity = m * g
where m is the mass of the block and g is the acceleration due to gravity.
The mass of the block can be calculated using its density and volume:
m = ρ * V = ρ * A * h
where ρ is the density of the block, A is the cross-sectional area of the block, and h is the height of the block.
The height of the block can be calculated as:
h = m / (ρ * A) = 4.58 kg / (625 kg/m^3 * A)
The cross-sectional area of the block is not given, so we cannot calculate it directly. However, we can use the fact that the block is in equilibrium to find the elongation of the spring.
When the block is in equilibrium, the force exerted by the spring is balanced by the weight of the block. The elongation of the spring (∆L) can be calculated using Hooke's Law:
F_spring = k * ∆L
where k is the spring constant.
Substituting the values we have:
F_gravity = F_spring
m * g = k * ∆L
Solving for ∆L:
∆L = (m * g) / k
We can calculate the spring constant using the given value of the spring's constant k:
k = 140 N/m
Substituting the values we have:
∆L = (m * g) / k = (4.58 kg * 9.8 m/s^2) / 140 N/m
∆L = 0.3186 m
Therefore, the elongation of the spring is 0.3186 m.