The length of a spring increases by 7.2 cm from its relaxed length when a mass of 1.4 kg is hanging in equilibrium from the spring. (a) What is the spring constant? (b) How much elastic potential energy is stored in the spring? (c) A different mass is suspended and the spring length increases by 12.2 cm from its relaxed length to its new equilibrium position. What is the second mass?
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The definition of spring constant (k) is force divided by deflection.
Elastic energy is (1/2)kX^2
Deflection is proportional to mass.
If you are not going to attempt problems like this, you are not going to learn physics.
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To solve this problem, we can use Hooke's Law, which states that the force exerted by a spring is directly proportional to its displacement from the equilibrium position.
(a) To find the spring constant, we can use the equation:
F = kx
Where F is the force exerted by the spring, k is the spring constant, and x is the displacement from the equilibrium position.
In this case, we know that the displacement is 7.2 cm and the mass is 1.4 kg. The weight of the mass is given by:
Weight = mass * gravitational acceleration
Weight = 1.4 kg * 9.8 m/s^2 = 13.72 N
Since the force exerted by the spring is equal to the weight of the mass at equilibrium, we have:
13.72 N = k * 0.072 m
Solving for k:
k = 13.72 N / 0.072 m = 190.56 N/m
Therefore, the spring constant is 190.56 N/m.
(b) To find the elastic potential energy stored in the spring, we can use the formula:
Elastic Potential Energy = 0.5 * k * x^2
Where k is the spring constant and x is the displacement from the equilibrium position.
In this case, the displacement is 7.2 cm, which is equal to 0.072 m. Plugging in the values, we have:
Elastic Potential Energy = 0.5 * 190.56 N/m * (0.072 m)^2
Elastic Potential Energy = 0.5 * 190.56 N/m * 0.005184 m^2
Elastic Potential Energy = 0.50266 J
Therefore, the elastic potential energy stored in the spring is approximately 0.50266 J.
(c) To find the second mass, we can use the same formula:
F = kx
Where F is the force exerted by the spring, k is the spring constant, and x is the displacement from the equilibrium position.
In this case, the displacement is 12.2 cm, which is equal to 0.122 m. Plugging in the values, we have:
F = k * 0.122 m
Since the force exerted by the spring is equal to the weight of the mass at equilibrium, we have:
F = second mass * gravitational acceleration
Plugging in the known values of the gravitational acceleration (9.8 m/s^2) and solving for the second mass:
k * 0.122 m = (second mass) * 9.8 m/s^2
190.56 N/m * 0.122 m = (second mass) * 9.8 m/s^2
23.23392 N = (second mass) * 9.8 m/s^2
(second mass) = 23.23392 N / 9.8 m/s^2
(second mass) = 2.37 kg
Therefore, the second mass is approximately 2.37 kg.
To answer these questions, we need to use Hooke's Law and the formula for elastic potential energy. Here are the steps to find the answers:
(a) What is the spring constant?
The spring constant, denoted by 'k', relates the force exerted by a spring to the displacement of the spring from its equilibrium position. Hooke's Law states that the force exerted by a spring is directly proportional to its displacement.
The formula for Hooke's Law is:
F = -kx
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 spring length increases by 7.2 cm, which is the displacement of the spring. The mass hanging in equilibrium creates an opposing force due to gravity.
The weight of the mass (force) exerted by gravity is given by:
F = mg
where m is the mass and g is the acceleration due to gravity (approximately 9.8 m/s^2).
Since the force exerted by the spring and the force exerted by gravity are equal in equilibrium, we can set up an equation using Hooke's Law and the weight of the mass:
-kx = mg
Rearranging the equation, we get:
k = -mg/x
Substituting the given values:
m = 1.4 kg
g = 9.8 m/s^2
x = 7.2 cm = 0.072 m
Calculating the spring constant, we have:
k = -(1.4 kg * 9.8 m/s^2) / 0.072 m
Simplifying the equation, we find:
k ≈ -192.22 N/m
Note: Since the displacement is in the positive direction (the spring length increases), the negative sign indicates that the force exerted by the spring is in the opposite direction (opposing the displacement).
Therefore, the spring constant is approximately 192.22 N/m.
(b) How much elastic potential energy is stored in the spring?
The elastic potential energy stored in a spring is the work done to stretch or compress it from its relaxed length.
The formula for elastic potential energy is:
PE = 0.5 * k * x^2
where PE is the elastic potential energy, k is the spring constant, and x is the displacement of the spring.
Substituting the given values:
k = 192.22 N/m (from part a)
x = 0.072 m
Calculating the elastic potential energy, we have:
PE = 0.5 * 192.22 N/m * (0.072 m)^2
Simplifying the equation, we find:
PE ≈ 0.5 * 192.22 N/m * 0.005184 m^2
PE ≈ 0.502 J (to 3 significant figures)
Therefore, the elastic potential energy stored in the spring is approximately 0.502 J.
(c) What is the second mass?
To find the second mass, we can again use Hooke's Law. Since the spring constant (k) remains the same, we can use the following equation:
F = -kx
In this case, the spring length increases by 12.2 cm, which is the displacement of the spring. We need to determine the new force exerted by the spring when the new mass hangs in equilibrium.
Setting up the equation, we have:
-kx = mg
Rearranging the equation, we get:
m = -kx / g
Substituting the given values:
k = 192.22 N/m (from part a)
x = 12.2 cm = 0.122 m
g = 9.8 m/s^2
Calculating the mass, we have:
m = -(192.22 N/m * 0.122 m) / 9.8 m/s^2
Simplifying the equation, we find:
m ≈ -2.4 kg
Since mass cannot be negative, we disregard the negative sign. Therefore, the second mass is approximately 2.4 kg.
Therefore, the second mass is approximately 2.4 kg.