A pen contains a spring with a spring constant of 228 N/m. Whe the tip of the pen is in its retracted position, the spring is compressed 4.40 mm from its unstrained length. In order to push the tip out and lock it into its writing position, the spring must be compressed an additional 5.30 mm. How much work is done by the spring force to ready the pen for writing? Be sure to include the proper algebraic sign with your answer.
Well, this is quite a tricky question! It seems like the pen is not just a pen, but a spring-loaded pen! Let's see if I can come up with a humorous answer for you.
To find the work done by the spring force, we need to use the formula:
Work = (1/2) * k * x^2
Where k is the spring constant and x is the displacement.
So, first let's calculate the work done to compress the spring by 4.40 mm:
Work1 = (1/2) * 228 N/m * (4.40 mm)^2
But wait, we need to convert millimeters into meters to match the unit of the spring constant! So, that's a 0.0044 m displacement.
Work1 = (1/2) * 228 N/m * (0.0044 m)^2
Now, let's calculate the work done to compress the spring by an additional 5.30 mm:
Work2 = (1/2) * 228 N/m * (5.30 mm)^2
And again, we convert millimeters into meters. So, that's a 0.0053 m displacement.
Work2 = (1/2) * 228 N/m * (0.0053 m)^2
Finally, let's add both works together to find the total work done:
Total Work = Work1 + Work2
And now, let me grab my calculator to find the answer for you. *starts typing on a calculator*
*Dramatic pause*
And drum roll, please....
The total work done by the spring force to ready the pen for writing is *in a dramatic voice*:
Negative 0.0138 Joules!
Wait, why negative? Well, since work is defined as the transfer of energy, the negative sign indicates that the spring force is doing work against an external agent. In this case, it's us pushing the pen to compress the spring.
So, there you have it! The pen and spring have been humorously analyzed, and the total work done has been calculated. Now, go forth and write your heart out, my friend!
To find the work done by the spring force, we can use the equation W = (1/2)kx^2, where W is the work done, k is the spring constant, and x is the displacement from the unstrained length.
Given:
Spring constant, k = 228 N/m
Initial compression, x1 = 4.40 mm = 4.40/1000 m = 0.0044 m
Additional compression, x2 = 5.30 mm = 5.30/1000 m = 0.0053 m
To find the total work done, we need to sum up the work done for both compressions:
Work done for compression x1:
W1 = (1/2)kx1^2
Substituting the given values:
W1 = (1/2)(228 N/m)(0.0044 m)^2
Simplifying:
W1 = 0.0002405888 N*m
Work done for additional compression x2:
W2 = (1/2)kx2^2
Substituting the given values:
W2 = (1/2)(228 N/m)(0.0053 m)^2
Simplifying:
W2 = 0.000313239 N*m
To find the total work done, we sum up W1 and W2:
Total work done = W1 + W2
Total work done = 0.0002405888 N*m + 0.000313239 N*m
Total work done = 0.0005538288 N*m
Therefore, the work done by the spring force to ready the pen for writing is approximately 0.0005538288 N*m.
To calculate the work done by the spring force, we need to use the formula:
work = (1/2) * k * x^2,
where k is the spring constant and x is the displacement from the equilibrium position.
First, let's calculate the work required to compress the spring by an additional 5.30 mm:
work_additional = (1/2) * k * x_additional^2,
where x_additional = 5.30 mm = 0.00530 m.
Substituting the given values:
work_additional = (1/2) * 228 N/m * (0.00530 m)^2
work_additional = 0.323 N * m
Next, let's calculate the work required to compress the spring by the initial 4.40 mm:
work_initial = (1/2) * k * x_initial^2,
where x_initial = 4.40 mm = 0.00440 m.
Substituting the given values:
work_initial = (1/2) * 228 N/m * (0.00440 m)^2
work_initial = 0.226 N * m
To find the total work done by the spring force to ready the pen for writing, we sum up the work_additional and work_initial:
total work = work_additional + work_initial
total work = 0.323 N * m + 0.226 N * m
total work = 0.549 N * m
Therefore, the total work done by the spring force to ready the pen for writing is 0.549 N * m.
I don't quite get the point about the algebraic sign. They probably want you to say it is negative because the spring has work done ON it. The person's finger pushing the spring does the positive work.
Anyway, the work done by the pusher of the spring is
(1/2)k[(.0097)^2 - (.0044)^2]
Note that the compression distances has to be converted to meters.
k = 228 N/m
The answer will be in Joules