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?
a. k=F/(L2-L1)=2*9.8/(15-10) = 3.92N/cm
= 392N/m.
b. F = 3.92N/cm * (20-10)cm = 39.2 N.
a = F/M = 39.2/2 = 19.6 m/s^2.
c. V^2 = Vo^2 + 2a*d = 0 + 2*19.6*0.10 =
3.92.
V = 1.98 m/s.
d.
(a) To find the stiffness (spring constant) of the spring, we 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. Mathematically, this can be expressed as:
F = k * x
Where F is the force exerted by the spring, k is the spring constant, and x is the displacement of the spring.
In this case, the mass of 2.0 kg is hung from the spring, causing the spring to stretch from its unstretched length of 10.0 cm to a length of 15 cm. The displacement of the spring can be calculated as:
x = 15 cm - 10.0 cm = 5.0 cm = 0.05 m (converting cm to m)
The weight of the mass can be calculated as:
W = m * g
Where W is the weight, m is the mass, and g is the acceleration due to gravity (approximately 9.8 m/s^2).
Substituting the values, we have:
W = 2.0 kg * 9.8 m/s^2 = 19.6 N
Since the spring is in equilibrium, the force exerted by the spring (F) is equal to the weight (W). Therefore:
F = 19.6 N
Using Hooke's Law, we can rearrange the formula to solve for k:
k = F / x
Substituting the values, we have:
k = 19.6 N / 0.05 m = 392 N/m (Newtons per meter)
The stiffness (spring constant) of the spring is 392 N/m.
(b) To determine the acceleration of the mass at the instant it is released, we can use Newton's second law of motion, which states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. Mathematically, this can be expressed as:
F = m * a
Where F is the net force, m is the mass, and a is the acceleration.
In this case, the net force acting on the mass is the force exerted by the spring when it is stretched from 20.0 cm to its equilibrium length. The displacement of the spring can be calculated as:
x = 20.0 cm - 10.0 cm = 10.0 cm = 0.10 m (converting cm to m)
Using Hooke's Law, we can calculate the force exerted by the spring:
F = k * x
Substituting the values, we have:
F = 392 N/m * 0.10 m = 39.2 N
Since the net force acting on the mass is in the opposite direction to its displacement, we have:
F = -39.2 N
Substituting the mass (m = 2.0 kg) into Newton's second law, we can solve for the acceleration (a):
-39.2 N = 2.0 kg * a
a = -39.2 N / 2.0 kg
a ≈ -19.6 m/s^2
The acceleration of the mass at the instant it is released is approximately -19.6 m/s^2, where the negative sign indicates that the acceleration is in the opposite direction of the displacement.
(c) If the mass accelerated at a constant rate, it would continue to accelerate until it reaches the equilibrium length of the spring. At this point, the net force on the mass would be zero, and it would decelerate until it comes to rest. The reason this is not actually what happens is due to the presence of additional forces acting on the mass, such as air resistance and friction. These forces oppose the motion of the mass and cause it to gradually lose energy, which results in it coming to rest before reaching the equilibrium length of the spring.
(d) When the mass reaches the equilibrium length of the spring, it is not moving. Therefore, the speed of the mass at this point is zero.
To solve this problem, we need to apply 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 formula for Hooke's Law is:
F = -kx
where F is the force exerted by the spring, k is the spring constant (or stiffness), and x is the displacement from the equilibrium position.
(a) To find the stiffness of the spring, we can use the given information that the spring has an unstretched length of 10.0 cm and becomes 15 cm long when the mass comes to rest. From this, we can calculate the displacement of the spring:
x = 15 cm - 10 cm = 5 cm = 0.05 m
Next, we need to determine the force exerted by the spring when the mass is at rest. The force exerted by the spring is equal in magnitude but opposite in direction to the force exerted by the mass due to gravity. So, we have:
F = mg
where m is the mass (2.0 kg) and g is the acceleration due to gravity (9.8 m/s^2).
F = (2.0 kg)(9.8 m/s^2) = 19.6 N
Now, we can rearrange Hooke's Law to solve for the spring constant:
k = -F / x
k = -19.6 N / 0.05 m = -392 N/m
The negative sign indicates that the force exerted by the spring is in the opposite direction of the displacement.
So, the stiffness (spring constant) of the spring is 392 N/m.
(b) To find the acceleration of the mass when it is released from a length of 20.0 cm, we can use Newton's second law of motion:
F = ma
The force exerted by the spring is given by Hooke's Law:
F = -kx
where k is the spring constant and x is the displacement.
At the equilibrium position, the displacement of the spring is 0. So, when the mass is released from a length of 20.0 cm, the displacement is:
x = 20.0 cm - 10.0 cm = 10.0 cm = 0.1 m
Substituting these values into the equation, we have:
-392 N/m * 0.1 m = m a
Simplifying this equation, we find:
a = (-392 N/m * 0.1 m) / 2.0 kg
a = -19.6 m/s^2
The negative sign indicates that the acceleration is in the opposite direction of the displacement.
Therefore, the acceleration of the mass at the instant when it is released is 19.6 m/s^2 in the opposite direction.
(c) If the mass accelerated at a constant rate, it would continue to accelerate and reach a higher velocity when it reaches the equilibrium length of the spring. However, this is not what actually happens in real-life situations due to the effects of air resistance and other external forces. These forces act on the mass and cause it to gradually slow down until it comes to rest at the equilibrium position.
(d) When the mass reaches the equilibrium length of the spring, it comes to rest and has zero velocity. Therefore, the mass is not moving when it reaches the equilibrium position.