A light spring of constant k = 156 N/m rests vertically on the bottom of a large beaker of water (Fig. P9.34a). A 4.10 kg block of wood (density = 650 kg/m3) is connected to the spring and the block-spring system is allowed to come to static equilibrium (Fig. P9.34b). What is the elongation, ΔL, of the spring?
See your 12-4-11,4:04pm post for solution
To find the elongation, ΔL, of the spring, we need to use Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement from its equilibrium position. In formula form, Hooke's Law is expressed as:
F = -k * ΔL
Where:
F is the force applied to the spring,
k is the spring constant, and
ΔL is the elongation or displacement of the spring.
In this problem, we are given the spring constant, k = 156 N/m. We need to calculate the force acting on the spring, F, which is equal to the weight of the block.
To calculate the weight of the block, we use the formula:
Weight = mass * gravity
Given that the density of the block, ρ = 650 kg/m3, and the mass of the block, m = 4.10 kg, we can find the volume of the block by dividing the mass by the density:
Volume = mass / density = 4.10 kg / 650 kg/m3 = 0.00631 m3
The weight of the block is then:
Weight = mass * gravity = 4.10 kg * 9.8 m/s2 = 40.18 N
Since the block is in static equilibrium, the force exerted by the spring is equal in magnitude but opposite in direction to the weight of the block. Thus, the force applied to the spring is -40.18 N.
Now we can use Hooke's Law to find the elongation, ΔL:
F = -k * ΔL
-40.18 N = -156 N/m * ΔL
Solving the equation for ΔL:
ΔL = (-40.18 N) / (-156 N/m)
ΔL ≈ 0.257 m
Therefore, the elongation of the spring, ΔL, is approximately 0.257 meters.