Find the Work done

Question

A force of  newtons is required to hold a spring that has been stretched from its natural length of 20 cm to a length of 30 cm. Find the work done in stretching the spring from 30 cm to 45 cm. First, setup an integral and find , , and  which would compute the amount of work done.



Answer 1

Hi Tuyen,

My name is Andrew, for this problem I would approach this by first drawing a picture of the diagram when the forces are applied. The first force is the one given where it equals 20 Newtons. All forces have to have an opposite force to counteract this and what is acting against the block is the spring itself.

Assumptions:

1. The spring is sitting on a table or desk attached to a fixed wall or something similar
2. the normal force is not acting against the block, otherwise a coefficient of friction would be given.

Equations to know:

1. Force of a spring: F_s = -k x
2. where k is the spring constant and x is the linear relationship of a spring
3. Work of a spring = W_s = ∫ F_s dx where the bounds are where the spring starts to where the spring ends.

so I would start by summing up the forces on the stretched spring. I cannot depict a picture here so I will describe the image. I drew my spring stretching to the right, so the force F_applied is to the right and then the F_s is to the left. ∑Forces you get 0 = F_applied - F_s . Now we can equate that the two forces are the same, this will come in handy in a moment.

My next calculation was to determine the work done on the spring. since the F_s and F_applied are the same we can place the work done on spring as ƒ

W = ∫ F dx

W = ∫ (20 Newtons) dx

The lower bound is where we started and at rest this is 0 to where the spring is at rest and the spring moves to 10 more cm to 30 cm. the next step is to change the cm to meters which is 0.1 meter, so the bound of integration are going to be from 0 to 0.1 meter.

W = 20x ... from 0 to 0.1

W = 20 (0.1 - 0)

W = 2 Joules

now we can solve for the k value by putting in the F_s equation into the integral instead so, W = ∫kx dx, and I think that here because the spring is being stretched the k value is positive. The bounds here are from 0 to 0.1 m as well. This is where we can relate the work on the spring to the work done by pulling the spring

W = 2 Joules

2 = (1/2)*k*x^2 from 0 to 0.1 meters

2 = (k/2) * (0.1^2 - 0)

4 = 0.01*k

k= 400 Newtons/meter

now to answer the question:

we are extending the spring again so k is positive.

W = ∫ kx dx from 30cm to 45cm

but not really. we are going from 10cm to 25cm or from 0.1 m to 0.25 m

W = ∫400*x dx from 0.1 to 0.25 m

W = (400/2) * x²

W = 200 ( 0.25² - 0.1²)

W = 10.5 Joules

The interesting thing to note here is that the work done to move the spring an extra 15 cm the work needed to do this is over 5 times the work to move the spring 10 cm. At this point when you get an answer you have to think about this answer and ask does this make sense, and if you have handled a spring this effect can be seen where the further you stretch a spring the "harder" it is to stretch. where "harder" usually a reference to how much work is being done.

to be clear I do not know what a and b are referring to in the initial question, but the total work done to move the spring to the location of 45 cm is 12.5 Joules.