Orientation on Calculus Work Problem. Hooke's Law.

In summary: The x in your integral is the extent of compression. The spring starts at its relaxed length of 13 inches. That's x = 0, no compression. It is compressed from 13 inches to 8 inches. That's a compression of 5 inches, x = 5. What is the value of x when the spring has been compressed by only the first of those five inches of compression?x = 5
  • #1
Dan350
44
0
A force of 16,000lb compresses a string from its natural length of 13 inch to 8 inch. Find the work done to compress it to the first inch

W=$$\int F dx$$


F=kx
16000=K(5)
3200=KW=$$\int F dx$$

W=$$\int_1^{13}\!\ 3200xdx$$

[1600x^2] from 1 to 13

w= 268800ftlb

Am I right?
I think the trick here is the limits.
If they would be from 0 to 1,, aint that to little work?
I mean, as you compress down, it's harder to do it. The work has to increase as you reach 0 inch
Am I right?
If not, please explain

Or is it from 12 to 13?

how I viewed is that they want to know the work as you compress it to the first inch.. in this case 1

I need a little orientation

| 16000Lb |
/----*------------------------*----------------------/
0inch 1inch 8inch 13inc
Thanks!

 
Last edited:
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  • #2
Dan350 said:
A force of 16,000ft-lb
Wrong units for a force.
160000=K(5)
Extra 0 crept in. What units do you want the work in? What units is the 5 in?
W=$$\int_1^{13}\!\ 3200xdx$$
What are the initial and final lengths of the spring as it goes through "the first inch" of its compression from 13 inches to 8 inches?
 
  • #3
haruspex said:
Wrong units for a force.

Extra 0 crept in. What units do you want the work in? What units is the 5 in?

What are the initial and final lengths of the spring as it goes through "the first inch" of its compression from 13 inches to 8 inches?

Sorry about that
force is 16000lb
using hooke's law I got the K
16000lb=K(5) 5 as the inches that the Force compressed

The length of the string is 13 inches
It was compressed 5 inches by a 16000lb force
I need to find the work of the first inch

Thanks
 
  • #4
Dan350 said:
The length of the string is 13 inches
A spring, presumably, not a string.
It was compressed 5 inches by a 16000lb force
I need to find the work of the first inch
So answer my question:
"What are the initial and final lengths of the spring as it goes through "the first inch" of its compression from 13 inches to 8 inches? "​
 
  • #5
haruspex said:
A spring, presumably, not a string.

So answer my question:
"What are the initial and final lengths of the spring as it goes through "the first inch" of its compression from 13 inches to 8 inches? "​


1 and 13
the spring is 13 inch long. from 0 to 13 there's no work, now, whether it compresses or streches, a work can be calculated
 
  • #6
Dan350 said:
1 and 13
No.
The x in your integral is the extent of compression. The spring starts at its relaxed length of 13 inches. That's x = 0, no compression. It is compressed from 13 inches to 8 inches. That's a compression of 5 inches, x = 5. What is the value of x when the spring has been compressed by only the first of those five inches of compression?
 

1. What is Hooke's Law?

Hooke's Law is a principle in physics that states the force required to stretch or compress a spring is directly proportional to the displacement of the spring from its equilibrium position.

2. How is Hooke's Law related to calculus?

Hooke's Law can be represented by a linear function in calculus, where the slope of the line represents the spring constant (k) and the x-intercept represents the equilibrium position of the spring. This allows us to use calculus concepts to solve problems involving Hooke's Law.

3. What is the equation for Hooke's Law?

The equation for Hooke's Law is F = -kx, where F is the force applied to the spring, k is the spring constant, and x is the displacement of the spring from its equilibrium position.

4. How do you solve calculus work problems involving Hooke's Law?

To solve calculus work problems involving Hooke's Law, you first need to set up the integral using the equation F = -kx. From there, you can use calculus techniques such as integration and differentiation to find the work done on the spring or the displacement of the spring.

5. Can Hooke's Law be applied to other systems besides springs?

Yes, Hooke's Law can be applied to other systems that exhibit linear elasticity, such as rubber bands and some types of metals. However, the equation may need to be modified to account for different variables and constants in these systems.

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