A certain amount of work is required to stretch a spring from its equilibrium position. (a) If twice the work is performed on the spring, the spring will stretch more by a factor of (1)square root of 2 (2)2,(3)1/square root of 2, (4)1/2. Why? (b) If 100J of work is done to pull a spring 1.0cm, what work is required to stretch it 3.0cm?
(a) correct answer is (1) square root of 2. but (b) is what i am having trouble with.
1/2 k 1^2 = 100
1/2 k 3^2 = ?
Thank you!
To answer part (b), we can make use of Hooke's Law, which states that the force exerted by a spring is directly proportional to its displacement from equilibrium.
Hooke's Law can be written as F = -kx, where F is the force exerted by the spring, k is the spring constant, and x is the displacement from equilibrium.
We can link the work done on the spring to the force and displacement using the formula for work:
Work = Force * Displacement * cos(theta)
Since the force and displacement are in the same direction (and cos(theta) = 1), we can write:
Work = Force * Displacement
From Hooke's Law, we know that the force exerted by the spring is F = -kx, so we can substitute this into the work equation:
Work = (-kx) * Displacement
Now, we can solve for the work required to stretch the spring from 1.0 cm to 3.0 cm. Let's assume the spring constant, k, is a constant value.
Work1 = (-k * 1.0 cm) * 1.0 cm = -k J
Work2 = (-k * 3.0 cm) * 3.0 cm = -9k J
To find the difference in work between stretching the spring from 1.0 cm to 3.0 cm, we can subtract Work1 from Work2:
ΔWork = Work2 - Work1 = (-9k J) - (-k J) = -8k J
Therefore, the work required to stretch the spring from 1.0 cm to 3.0 cm is -8k J.
Please note that the negative sign indicates that work is done on the spring, which results in potential energy being stored in the spring.
To solve part (b) of the problem, we can use the concept of elastic potential energy stored in a spring.
The work done to stretch a spring is equal to the change in potential energy of the spring. The potential energy stored in a spring can be calculated using the formula:
Elastic Potential Energy = (1/2) * k * x^2
Where:
- k is the spring constant (a property of the spring)
- x is the displacement or stretch of the spring from its equilibrium position.
In this case, we know that 100J of work is done to pull the spring 1.0cm. We need to find the work required to stretch it 3.0cm (x = 3.0cm = 0.03m).
First, let's find the spring constant, k. Since we don't have the specific information about the spring, we can't directly calculate the spring constant. However, we can use Hooke's Law to find it indirectly.
Hooke's Law states that the force required to stretch or compress a spring is proportional to the displacement, x. Mathematically, it can be expressed as:
F = -k * x
Where:
- F is the force applied to the spring.
Since we know that 100J of work is done to stretch the spring by 1.0cm, we can calculate the force applied using the work-energy principle:
Work Done = Force * Distance
100J = F * 0.01m
Solving for force:
F = 100J / 0.01m
F = 10,000N
Now, we can use Hooke's Law to find the spring constant:
10,000N = -k * 0.01m
Solving for k:
k = 10,000N / 0.01m
k = 1,000,000 N/m
Now that we have the spring constant, we can calculate the work required to stretch the spring by 3.0cm.
Elastic Potential Energy = (1/2) * k * x^2
Potential Energy = (1/2) * 1,000,000 N/m * (0.03m)^2
Potential Energy = (1/2) * 1,000,000 N/m * 0.0009m^2
Potential Energy = 450 J
Therefore, the work required to stretch the spring by 3.0cm is 450 J.