A block with mass of 5.0kg is suspended from an ideal spring having negligible mass and stretches the spring 0.20m to its equilibrium position.
A) What is the force constant of the spring?
B) The spring is then stretched 0.80m from its equilibrium position and then released with the velocity zero. Calculate the velocity of the mass when the mass passes again through the equilibrium position.
A) k = m g / x = 5.0 * 9.8 / .20 N/m
B) find the work done (energy) in stretching the spring from .20 m to .80 m
... 1/2 k x^2 ... for each position
the work is energy stored in the spring
... when the mass is released, the stored energy accelerates it upward
... the stored energy, minus the change in position (gravitational energy)
is the kinetic energy of the mass at the equilibrium point
A) To find the force constant of the spring, we can use Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement from its equilibrium position.
The force exerted by the spring can be calculated using the equation: F = k * x
Where:
F = Force exerted by the spring
k = Force constant (also known as spring constant)
x = displacement from the equilibrium position
From the given information, we know that the mass of the block is 5.0 kg and the spring is stretched 0.20 m.
Let's substitute the values into the equation to find the force:
F = k * x
F = k * 0.20 m
F = k * 0.20 N
Now, let's calculate the force constant (k):
Rearranging the equation, we have:
k = F / x
k = (5.0 kg * 9.8 m/s^2) / 0.20 m
Using a calculator, we can calculate the force constant:
k = 245 N/m
Therefore, the force constant of the spring is 245 N/m.
B) To calculate the velocity of the mass when it passes through the equilibrium position again, we need to consider the conservation of mechanical energy.
The potential energy stored in the spring when it is stretched is given by the equation: PE = (1/2) * k * x^2
The block has zero velocity when it passes through the equilibrium position, which means all the initial potential energy is converted into kinetic energy.
The kinetic energy of the block when it passes through the equilibrium position can be calculated using the equation: KE = (1/2) * m * v^2
Setting the potential energy equal to the kinetic energy, we have:
(1/2) * k * x^2 = (1/2) * m * v^2
Rearranging the equation to solve for velocity (v), we get:
v^2 = (k * x^2) / m
Taking the square root of both sides, we find:
v = √((k * x^2) / m)
Substituting the known values into the equation, we have:
v = √((245 N/m * (0.80 m)^2) / 5.0 kg)
Using a calculator, we can calculate the velocity:
v ≈ 4.43 m/s
Therefore, the velocity of the mass when it passes through the equilibrium position again is approximately 4.43 m/s.
To find the force constant of the spring (question A), we can use Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement from its equilibrium position. The formula for Hooke's Law is:
F = -kx
Where:
F = Force exerted by the spring
k = Force constant of the spring
x = Displacement from the equilibrium position
Given:
Mass of the block (m) = 5.0 kg
Displacement from the equilibrium position (x) = 0.20 m
We know that force (F) can be calculated using Newton's second law:
F = m * a
Where:
m = Mass of the block
a = Acceleration of the block
Since the block is in equilibrium (velocity is zero), the net force acting on it is zero. Therefore, the force provided by the spring balances the force due to gravity:
F_spring = F_gravity
Using Newton's second law, we can write:
kx = mg
Solving for the force constant (k):
k = mg / x
Substituting the given values:
k = (5.0 kg) * (9.8 m/s²) / (0.20 m)
Now, let's calculate the value of k:
k = 245 N/m
Therefore, the force constant of the spring is 245 N/m.
Moving on to question B, we are asked to find the velocity of the mass when it passes through the equilibrium position again. To do this, we need to use the principle of conservation of mechanical energy.
When the spring is stretched further to a displacement of 0.80 m from its equilibrium position and then released, it will oscillate between the two extreme positions. At the extreme points, all the potential energy is converted into kinetic energy, which can be used to calculate the velocity.
The formula for the potential energy stored in a spring is:
PE = 1/2 * k * x²
Where:
PE = Potential energy stored in the spring
k = Force constant of the spring
x = Displacement from the equilibrium position
At the maximum displacement (0.80 m), all the potential energy is converted into kinetic energy:
PE_max = KE_max
Using the formula for kinetic energy:
KE = 1/2 * m * v²
Where:
KE = Kinetic energy of the block
m = Mass of the block
v = Velocity of the block
Equating the potential energy to the kinetic energy:
1/2 * k * x² = 1/2 * m * v²
Simplifying the equation:
k * x² = m * v²
v² = (k * x²) / m
Taking the square root of both sides:
v = √((k * x²) / m)
Substituting the given values:
v = √((245 N/m) * (0.80 m)²) / (5.0 kg)
Now, let's calculate the value of v:
v ≈ 2.21 m/s
Therefore, the velocity of the mass when it passes through the equilibrium position again is approximately 2.21 m/s.