A light spring of constant k = 158 N/m rests vertically on the bottom of a large beaker of water (Fig. P9.34a). A 5.55 kg block of wood (density = 650 kg/m3) is connected to the spring and the mass-spring system is allowed to come to static equilibrium (Fig. P9.34b). What is the elongation, ÄL, of the spring?
Wb = mg = 5.55kg * 9.8N/kg = 54.39N.
= Weight of block.
d = 54.39/158 = 0.344 m Elongated.
thats wrong (above)
To find the elongation of the spring, we need to determine the change in length of the spring from its rest position when the block of wood is added.
First, let's calculate the weight of the block of wood. The weight can be found using the formula:
Weight = Mass * Acceleration due to gravity
The mass of the block is given as 5.55 kg. The acceleration due to gravity is approximately 9.8 m/s^2. Therefore,
Weight = 5.55 kg * 9.8 m/s^2 = 54.39 N
Now, let's determine the buoyant force acting on the block of wood when it is submerged in water. The buoyant force is equal to the weight of the water displaced by the block. To find the displaced volume of water, we can use the density of the block and the volume it occupies.
The volume of the block can be calculated using its mass and density:
Volume = Mass / Density
Volume = 5.55 kg / 650 kg/m^3 = 0.00853 m^3
The buoyant force can be determined using the formula:
Buoyant force = Density of water * Volume of water displaced * Acceleration due to gravity
Let's assume the density of water is approximately 1000 kg/m^3. Therefore,
Buoyant force = 1000 kg/m^3 * 0.00853 m^3 * 9.8 m/s^2 = 83.99 N
Since the block is in static equilibrium, the force exerted by the spring must balance the difference between the block's weight and the buoyant force. Therefore, the elongation of the spring can be found using Hooke's Law:
F_spring = k * ÄL
where F_spring is the force exerted by the spring, k is the spring constant, and ÄL is the elongation of the spring.
The net force acting on the block is given by:
Net force = Weight - Buoyant force
Net force = 54.39 N - 83.99 N = -29.6 N
Since the force exerted by the spring is equal in magnitude but opposite in direction to the net force, we have:
F_spring = -(-29.6 N) = 29.6 N
Plugging in the values into Hooke's Law equation, we get:
29.6 N = 158 N/m * ÄL
Therefore, the elongation of the spring, ÄL, is given by:
ÄL = 29.6 N / 158 N/m = 0.187 m (or 18.7 cm)