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) To find the force constant of the spring, you need to use Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement of the spring from its equilibrium position. The equation for Hooke's Law is:
F = -kx
Where:
- F is the force exerted by the spring
- k is the force constant (also known as the spring constant)
- x is the displacement from the equilibrium position
In this case, the mass is suspended from the spring, so the force exerted by the spring is equal to the weight of the mass, which can be calculated using the equation:
F = mg
Where:
- m is the mass of the block (5.0 kg)
- g is the acceleration due to gravity (approximately 9.8 m/s^2)
Therefore, we can set up an equation using Hooke's Law:
mg = -kx
Substituting the given values:
(5.0 kg)(9.8 m/s^2) = -k(0.20 m)
Solving for k:
k = -(5.0 kg)(9.8 m/s^2) / 0.20 m
By calculating this expression, you will find the force constant of the spring.
B) To calculate the velocity of the mass when it passes again through the equilibrium position, you can use the principle of conservation of mechanical energy.
When the spring is stretched to a displacement of 0.80 m and released, it possesses potential energy, which gradually converts into kinetic energy as the mass moves. At the equilibrium position, all potential energy is converted into kinetic energy.
The gravitational potential energy when the spring is stretched to 0.80 m is given by:
PE = mgh
Where:
- m is the mass of the block (5.0 kg)
- g is the acceleration due to gravity (approximately 9.8 m/s^2)
- h is the displacement of the spring from its equilibrium position (0.80 m)
The total mechanical energy (E) of the system, consisting of potential energy (PE) and kinetic energy (KE), is conserved. Therefore:
E = PE + KE
At the equilibrium position, the potential energy is zero, so the total mechanical energy is equal to the kinetic energy:
E = KE
Using the equation for kinetic energy:
KE = (1/2)mv^2
Where:
- m is the mass of the block (5.0 kg)
- v is the velocity of the mass
Therefore:
KE = E = (1/2)(5.0 kg)v^2
Now, equate the potential energy to the kinetic energy:
mgh = (1/2)mv^2
Substituting the given values:
(5.0 kg)(9.8 m/s^2)(0.80 m) = (1/2)(5.0 kg)v^2
Simplifying the equation and solving for v, you can determine the velocity of the mass when it passes again through the equilibrium position.