This assignment is about energy. But one of the new things we’ve just learned about is spring forces, so there should be a question about them. So the first parts of this question are not about energy.
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) How fast is the mass actually going when it reaches the equilibrium length of the spring?
To solve this problem, we will need to apply principles of spring forces and energy.
(a) To find the stiffness or spring constant (k) of the spring, we can use Hooke's Law, which states that the force exerted by the spring is directly proportional to the displacement of the spring from its equilibrium position. Hooke's Law can be written as F = -kx, where F is the force, k is the spring constant, and x is the displacement.
In this case, let's assume the equilibrium position of the spring is at its unstretched length of 10.0 cm. When the 2.0 kg mass is hung from the spring, it stretches to a length of 15 cm. The displacement (x) of the spring is calculated as 15 cm - 10 cm = 5 cm = 0.05 m.
Now, we know that the force exerted by the spring is equal to the weight of the mass (mg), where m is the mass and g is the acceleration due to gravity (approximately 9.8 m/s^2). Therefore, we can equate the two equations:
mg = kx
Substituting the known values, we have:
(2.0 kg)(9.8 m/s^2) = k(0.05 m)
Simplifying:
19.6 N = 0.05 k
Dividing both sides by 0.05:
k = 19.6 N / 0.05 m = 392 N/m
Therefore, the stiffness or spring constant of the spring is 392 N/m.
(b) To find the acceleration of the mass when it is released from a 20.0 cm length, we will use the concept of potential energy. The potential energy stored in the spring is equal to (1/2)kx^2, where k is the spring constant and x is the displacement from the equilibrium position.
At a length of 20.0 cm, the displacement x is calculated as 20 cm - 10 cm = 10 cm = 0.10 m.
The potential energy stored in the spring can be calculated as:
PE = (1/2)kx^2
= (1/2)(392 N/m)(0.10 m)^2
= 1.96 J
The potential energy will be converted to kinetic energy when the mass is released. Therefore, the kinetic energy of the mass at the instant of release is equal to 1.96 J.
Using the equation for kinetic energy, KE = (1/2)mv^2, where m is the mass and v is the velocity, we can find the velocity (v) of the mass. Rearranging the equation, we have:
v = √(2KE / m)
= √(2 * 1.96 J / 2.0 kg)
= √0.98 m^2/s^2
≈ 0.99 m/s
Therefore, the acceleration of the mass at the instant when it is released is approximately 0.99 m/s^2.
(c) If the mass accelerated at a constant rate, it would continue to accelerate until it reaches the equilibrium length of the spring. However, this is not what happens in reality due to the presence of a restoring force provided by the spring. As the mass moves away from the equilibrium position, the spring exerts a force opposing the displacement, causing a deceleration.
(d) When the mass reaches the equilibrium length of the spring, it momentarily comes to rest before moving in the opposite direction. At this point, the mass has zero velocity. Therefore, the speed of the mass is actually zero when it reaches the equilibrium length of the spring.
(a) To find the stiffness (spring 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. Mathematically, this can be expressed as F = -kx, where F is the force, k is the spring constant, and x is the displacement.
In this case, when the mass is hanging at rest, the spring is stretched by 15 cm - 10 cm = 5 cm = 0.05 m. 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 and g is the acceleration due to gravity (approximately 9.8 m/s^2).
Therefore, the spring constant can be determined as follows:
F = -kx
mg = -k(0.05)
2.0 kg * 9.8 m/s^2 = -k(0.05 m)
k = (2.0 kg * 9.8 m/s^2) / 0.05 m
Calculating this, we find that the stiffness (spring constant) of the spring is k ≈ 392 N/m.
(b) To find the acceleration of the mass when it is released, we can use the equation F = ma, where F is the net force acting on the mass and a is its acceleration.
When the mass is released, the net force acting on it is the sum of the force due to gravity and the force exerted by the spring. This can be calculated as follows:
F_net = mg + (-kx)
F_net = 2.0 kg * 9.8 m/s^2 + (-392 N/m * 0.2 m)
F_net = 19.6 N - 78.4 N
F_net = -58.8 N
Since the mass is released and there is no external force acting on it, this net force will cause the mass to accelerate upwards.
Using F_net = ma, we can solve for the acceleration:
-58.8 N = 2.0 kg * a
a = (-58.8 N) / (2.0 kg)
Calculating this, we find that the acceleration of the mass at the instant when it is released is approximately a ≈ -29.4 m/s^2.
Note: The negative sign indicates that the acceleration is upwards, opposite to the direction of gravity.
(c) If the mass were to accelerate at a constant rate until it reached the equilibrium length of the spring, it would be subject to a constant net force. This is because the force exerted by the spring is proportional to the displacement from the equilibrium position.
However, in reality, the spring force follows Hooke's Law, which means that the force exerted by the spring increases as the spring stretches or compresses further from the equilibrium position. As the mass is displaced more, the spring force becomes stronger, leading to a non-constant net force on the mass. This means that the mass does not accelerate at a constant rate as it approaches the equilibrium length of the spring.
(d) When the mass reaches the equilibrium length of the spring, it means that the net force acting on it is zero. At this point, the downward force due to gravity is balanced by the upward force exerted by the spring.
In this situation, the mass is momentarily at rest (velocity = 0 m/s) as it passes through the equilibrium position. Therefore, its speed is 0 m/s when it reaches the equilibrium length of the spring.