One end of a spring is attached to the ceiling. The unstretched length of the spring is 10.0 cm. A 2.0 kg mass is hung from the other end of the spring. It is slowly lowered until it comes to rest. At this point the spring is 15 cm long.
(a) What is the stiffness (spring constant) of the spring?
(b) The mass is now pulled down until the length of the spring is 20.0 cm. It is released. What
is the acceleration of the mass at the instant when it is released?
(c) If the mass accelerated at a constant rate how fast would it be going when it reached the
equilibrium length of the spring? Explain why this is not actually what happens.
(d) (We now return you to your regularly scheduled assignment about energy). How fast is the
mass actually going when it reaches the equilibrium length of the spring?
Electrical
To solve this problem, we will use Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement from its equilibrium position.
(a) To find the stiffness or spring constant (k), we can use Hooke's law. The formula for Hooke's law is:
F = -kx
Where:
F is the force applied to the spring
k is the spring constant
x is the displacement from the equilibrium position
In this case, we have a mass of 2.0 kg hanging from the spring, so the force applied to the spring is given by:
F = mg
where m is the mass and g is the acceleration due to gravity (approximately 9.8 m/s²).
The displacement from the equilibrium position is given by:
x = 15 cm - 10 cm = 5 cm = 0.05 m
Now we can plug these values into Hooke's law:
mg = kx
Solving for k:
k = mg / x
Substituting the given values:
k = (2.0 kg)(9.8 m/s²) / 0.05 m
k = 392 N/m
So, the stiffness or spring constant of the spring is 392 N/m.
(b) When the mass is pulled down until the length of the spring is 20.0 cm and then released, it will oscillate up and down. To find the acceleration of the mass at the instant it is released, we can use the equation for simple harmonic motion:
a = -(k/m) * x
where a is the acceleration, k is the spring constant, m is the mass, and x is the displacement from the equilibrium position.
At the instant of release, the displacement from the equilibrium position is 20.0 cm - 10.0 cm = 10.0 cm = 0.1 m.
Plugging in the values:
a = -(392 N/m) * (0.1 m) / 2.0 kg
a = -1.96 m/s²
So, the acceleration of the mass at the instant of release is -1.96 m/s², indicating that it is moving in the opposite direction of the displacement.
(c) If the mass were to accelerate at a constant rate, it would continue to move without oscillating and would eventually reach the equilibrium length of the spring. However, this is not what actually happens because the force exerted by the spring is not constant. As the mass moves away from the equilibrium position, the force becomes stronger, pulling it back towards the equilibrium.
(d) To find the speed of the mass when it reaches the equilibrium length of the spring, we can use the principle of conservation of mechanical energy. In simple harmonic motion, the total mechanical energy is conserved and given by:
E = 0.5 * k * x² + 0.5 * m * v²
where E is the total mechanical energy, k is the spring constant, x is the displacement from the equilibrium position, m is the mass, and v is the velocity.
At the equilibrium length, the displacement is zero, so the first term in the equation becomes zero. The total mechanical energy only depends on the kinetic energy.
E = 0.5 * m * v²
Rearranging the equation:
v = sqrt(2E / m)
Substituting the given values:
v = sqrt(2 * (0.5 * k * x²) / m)
v = sqrt((k * x²) / m)
Plugging in the values:
v = sqrt((392 N/m * (0.10 m)²) / 2.0 kg)
v = sqrt(3.92 m²/s²)
v = 1.98 m/s
Therefore, the mass is actually moving with a speed of 1.98 m/s when it reaches the equilibrium length of the spring.