A spring with k = 54 N/cm is initially stretched 1 cm from its equilibrium length.
(a) How much more energy is needed to further stretch the spring to 2 cm beyond its equilibrium length?
(b) From this new position, how much energy is needed to compress the spring to 2 cm shorter than its equilibrium position?
To find the energy needed to stretch or compress a spring, we can use the formula for elastic potential energy:
Elastic potential energy = 0.5 * k * x^2
where k is the spring constant and x is the displacement from the equilibrium position.
(a) To find the energy needed to stretch the spring to 2 cm beyond its equilibrium position, we need to calculate the elastic potential energy at that displacement.
Given:
k = 54 N/cm (spring constant)
x = 2 cm (displacement)
Converting cm to meters:
x = 2 cm = 2/100 m = 0.02 m
Substituting the values into the formula:
Elastic potential energy = 0.5 * k * x^2
Elastic potential energy = 0.5 * 54 N/cm * (0.02 m)^2
Elastic potential energy = 0.5 * 54 N/cm * 0.0004 m^2
Elastic potential energy ≈ 0.000432 J
Therefore, approximately 0.000432 joules of energy are needed to further stretch the spring to 2 cm beyond its equilibrium length.
(b) To find the energy needed to compress the spring to 2 cm shorter than its equilibrium position, we need to calculate the elastic potential energy at that displacement.
Given:
k = 54 N/cm (spring constant)
x = -2 cm (displacement)
Converting cm to meters:
x = -2 cm = -2/100 m = -0.02 m
Substituting the values into the formula:
Elastic potential energy = 0.5 * k * x^2
Elastic potential energy = 0.5 * 54 N/cm * (-0.02 m)^2
Elastic potential energy = 0.5 * 54 N/cm * 0.0004 m^2
Elastic potential energy ≈ 0.000432 J
Therefore, approximately 0.000432 joules of energy are needed to compress the spring to 2 cm shorter than its equilibrium position.
To solve this problem, we need to understand the relationship between the force applied to the spring and the displacement from its equilibrium position. For a spring, this relationship is described by Hooke's Law, which states that the force exerted by a spring is directly proportional to its displacement.
Hooke's Law equation:
F = -kx
Where:
F = Force exerted by the spring
k = Spring constant (stiffness)
x = Displacement from the equilibrium position. Positive for stretching, negative for compression.
Now, let's solve the given problems:
(a) How much more energy is needed to further stretch the spring to 2 cm beyond its equilibrium length?
To find the additional energy needed to stretch the spring further, we can use the formula for the potential energy stored in a spring:
Potential Energy (PE) = (1/2) * k * x^2
The initial displacement of the spring is 1 cm, and we need to stretch it by an additional 2 cm. Therefore, the total displacement will be 1 cm + 2 cm = 3 cm.
Let's plug in these values into the formula:
PE = (1/2) * k * (3 cm)^2
PE = (1/2) * 54 N/cm * (3 cm)^2
PE = 243 N cm
Therefore, to stretch the spring by an additional 2 cm beyond its equilibrium length, 243 N cm of energy is needed.
(b) From this new position, how much energy is needed to compress the spring to 2 cm shorter than its equilibrium position?
Now, we need to find the potential energy required to compress the spring by 2 cm from the new position. Here, we'll consider the displacement as negative (-2 cm) since it represents compression.
Using the same formula for the potential energy stored in a spring:
PE = (1/2) * k * x^2
PE = (1/2) * 54 N/cm * (-2 cm)^2
PE = 108 N cm
Therefore, to compress the spring by 2 cm shorter than its equilibrium position from the new stretched position, 108 N cm of energy is needed.
PE1 =k•x1²/2
PE2 = k•x2²/2
W = ΔPE =PE2 –PE1 =k(x2² - x1²)/2 = ….
where k =54 N/m, x1 =0.01 m,
x2 =0.01+0.02 = 0.03 m.
this compression will be done by the work of elastic force of the spring. When we release the spring which is stretched by 0.03 m, the released energy is k•(0.003)²/2 which is large than k•(0.02)²/2.
I'm still a little confused. When I use the equation:
W = k(x2^2 - x1^2)/2, I get 0.0216 J. This doesn't seem correct.