A spring scale is calibrated from zero to 20. N. The calibrations extend over a length of 0.10 m. (a) What is the elastic potential energy of the spring in the scale when a weight of 5.0 N hangs from it? (b) What is the elastic potential energy when the spring is fully stretched?
(a) To find the elastic potential energy of the spring in the scale when a weight of 5.0 N hangs from it, we need to use the formula for elastic potential energy:
Elastic Potential Energy (PE) = 0.5 * k * x^2
Where:
- PE is the elastic potential energy
- k is the spring constant
- x is the displacement from the equilibrium position (the stretched length of the spring)
In this case, since the scale is calibrated from zero to 20 N over a length of 0.10 m, we can find the spring constant (k) using the formula:
k = (force / displacement)
Given force = 20 N and displacement = 0.10 m, we can calculate the spring constant:
k = 20 N / 0.10 m
k = 200 N/m
Now, we can calculate the elastic potential energy when a weight of 5.0 N hangs from it. The displacement (x) in this case would be the difference between the fully stretched length (0.10 m) and the equilibrium position (0 m):
Displacement (x) = 0.10 m - 0 m
x = 0.10 m
Substituting the values into the formula, we have:
PE = 0.5 * (200 N/m) * (0.10 m)^2
PE = 0.5 * 200 N/m * 0.01 m^2
PE = 0.5 * 2 N * m
PE = 1 J
Therefore, the elastic potential energy of the spring in the scale when a weight of 5.0 N hangs from it is 1 J.
(b) To find the elastic potential energy when the spring is fully stretched, we need to calculate the maximum displacement (x) of the spring.
Given that the fully stretched length of the spring is 0.10 m, the displacement (x) would be equal to the fully stretched length:
x = 0.10 m
Substituting this value into the formula for elastic potential energy:
PE = 0.5 * (200 N/m) * (0.10 m)^2
PE = 0.5 * 200 N/m * 0.01 m^2
PE = 0.5 * 2 N * m
PE = 1 J
Therefore, the elastic potential energy of the spring when it is fully stretched is also 1 J.
To calculate the elastic potential energy of a spring, we need to use the formula:
Elastic Potential Energy = ½ * k * x^2
Where:
- k is the spring constant (which determines the stiffness of the spring)
- x is the displacement of the spring from its equilibrium position
However, in order to use this formula, we need to determine the spring constant of the scale.
To find the spring constant (k), we need to use the information provided by the spring scale. It is mentioned that the spring scale is calibrated from 0 to 20 N over a length of 0.10 m.
The spring constant (k) can be found by using Hooke's Law, which states that the force applied to a spring is directly proportional to its displacement from its equilibrium position:
F = k * x
In this case, we can use the calibration information to determine the spring constant:
(20 N - 0 N) = k * (0.10 m - 0 m) --> 20 N = k * 0.10 m
Solving for k:
k = 20 N / 0.10 m
k = 200 N/m
Now that we know the spring constant (k), we can proceed to answer the given questions:
(a) What is the elastic potential energy of the spring in the scale when a weight of 5.0 N hangs from it?
Using the formula for elastic potential energy:
Elastic Potential Energy = ½ * k * x^2
For a weight of 5.0 N hanging from the spring scale, we need to find the displacement (x) of the spring. Displacement (x) can be determined by dividing the weight by the spring constant (k):
x = 5.0 N / 200 N/m
x = 0.025 m
Now we can calculate the elastic potential energy:
Elastic Potential Energy = ½ * (200 N/m) * (0.025 m)^2
(b) What is the elastic potential energy when the spring is fully stretched?
When the spring is fully stretched, it means it is displaced by its maximum value. In this case, the maximum displacement (x) would be 0.10 m since that is the length over which the spring scale is calibrated.
Using the formula for elastic potential energy:
Elastic Potential Energy = ½ * (200 N/m) * (0.10 m)^2
By plugging in the values, you can calculate the elastic potential energy when the spring is fully stretched.