Does it take twice the work to stretch a spring 2cm from its equilibrium position as it does to stretch it 1cm from its equilibrium position?
Yes it does.
(Just want to double check my answer)
To confirm the answer, let's explain how to calculate the work done in stretching a spring.
The work done to stretch a spring can be calculated using Hooke's law, which states that the force needed to stretch or compress a spring is directly proportional to the displacement from its equilibrium position. Hooke's law can be expressed by the equation:
F = -kx
where F is the force applied to the spring, k is the spring constant, and x is the displacement from the equilibrium position.
The work done (W) in stretching a spring is given by the equation:
W = (1/2)kx^2
where W is the work done.
Now, let's consider stretching the spring 1cm from its equilibrium position. So, x = 1cm = 0.01m.
Substituting the values into the equation, we get:
W1 = (1/2)k(0.01)^2
W1 = (1/2)k(0.0001)
W1 = 0.00005k
Similarly, let's consider stretching the spring 2cm from its equilibrium position. So, x = 2cm = 0.02m.
Substituting the values into the equation, we get:
W2 = (1/2)k(0.02)^2
W2 = (1/2)k(0.0004)
W2 = 0.0002k
Comparing the two work equations, we can see that W2 = 4W1, which means that stretching the spring 2cm requires four times the work of stretching it 1cm.
Therefore, it does not take twice the work to stretch a spring 2cm as it does to stretch it 1cm. Instead, it takes four times the work.