How Do Springs Help Support Truck Loads?

In summary, the problem involves trucks with leaf and helper springs attached to the axle. The leaf spring has a compression limit of 0.5m and a stiffness (k) of 5.25*10^5N/m, while the helper spring has a stiffness of 3.6*10^5N/m. For a load of 5.00*10^5N, the compression of the leaf spring is found to be 0.768m. To determine the work done in compressing the springs, the potential energy stored in each spring must be calculated separately. By solving for the displacement of each spring and summing the work done by both, the correct answer can be obtained.
  • #1
alexkolb
8
0
Problem statement:

Trucks have leaf and helper spring (spring put on top of leaf spring vertically) attached to axel. After the leaf spring is compressed by 0.5m, the helper spring helps with additional load.
k of leaf spring is 5.25*10^5N/m, and k of helper spring is 3.6*10^5N/m. Load is 5.00*10^5N

a) What is the compression of the leaf spring for the above load?

I solved by first finding what load leaf spring supports, since its max compression befor helper spring is activated is 0.5m, so i found F with x=0.5, and k=5.25*10^5. Then I subtracted that number from the total load to find how much do both spring support, and then used the sum of both k to find x when two springs are supporting the weight. then 0.5 plus that x is the compression of leaf spring, which gives 0.768m, and that's the answer in the book.

Now i have a problem with part b:

b) How much work is done in compressing the springs?

Any help would be appreciated, thanks!
 
Physics news on Phys.org
  • #2
alexkolb said:
Problem statement:

Trucks have leaf and helper spring (spring put on top of leaf spring vertically) attached to axel. After the leaf spring is compressed by 0.5m, the helper spring helps with additional load.
k of leaf spring is 5.25*10^5N/m, and k of helper spring is 3.6*10^5N/m. Load is 5.00*10^5N

a) What is the compression of the leaf spring for the above load?

I solved by first finding what load leaf spring supports, since its max compression befor helper spring is activated is 0.5m, so i found F with x=0.5, and k=5.25*10^5. Then I subtracted that number from the total load to find how much do both spring support, and then used the sum of both k to find x when two springs are supporting the weight. then 0.5 plus that x is the compression of leaf spring, which gives 0.768m, and that's the answer in the book.

Now i have a problem with part b:

b) How much work is done in compressing the springs?

Any help would be appreciated, thanks!
Hint: The work done to compress a spring is the potential energy stored in the spring.
 
  • #3
i did it that way

took k of 1st spring, times 0.5 squared, and over 2
then add that
to k of first and k of second spring, times displacement with 2 springs, 0.268 squared, over 2

it gives about 0.97*10^5J, which is not the ansewr that book gives...

book gives 1.68*10^5... i don't know how they get that
helpp...
 
Last edited:
  • #4
Each spring has a different displacement. You've actually solved it, you just don't know it yet. The leaf spring is compressed by 0.768 m, the helper spring is compressed by 0.268 m.

The total work done is the sum of the work done on/by these two springs.

Dorothy
 
  • #5
ok yeh that worked now...! :)

i messed up on it because i was adding Ks of both springs for second part of motion, and taking only one spring for first part;, in part a i added Ks so i didnt see why it wouldn't work in part, B, but you are right, its the displacement of each separate spring.

thanks!
 
  • #6
Ummm. I don't see why that would have worked... I would have solved it this way:

F = (k_leaf)(x_leaf) + (k_helper)(x_helper)
x_helper = x_leaf - 0.5m

And then solved that equation for x_leaf. Are you sure that you added the K's together?

Well, the right answer is always a good thing to have, in any case.

Good night,
Dorothy
 

1. What is the concept of work done by a spring?

The concept of work done by a spring refers to the amount of energy that is transferred when a spring is compressed or stretched. This energy is stored in the spring in the form of potential energy and is released when the spring returns to its original position.

2. How is work done by a spring calculated?

The work done by a spring can be calculated by multiplying the force applied to the spring by the distance the spring is compressed or stretched. This can be represented by the formula W = F * x, where W is the work done, F is the force applied, and x is the distance the spring is compressed or stretched.

3. Is the work done by a spring always positive?

No, the work done by a spring can be either positive or negative depending on the direction of the force applied. If the force is applied in the same direction as the displacement of the spring, the work done will be positive. However, if the force is applied in the opposite direction, the work done will be negative.

4. How does the spring constant affect the work done by a spring?

The spring constant, which is a measure of the stiffness of a spring, directly affects the work done by a spring. A higher spring constant means that the spring is stiffer and will require more force to compress or stretch it, resulting in a greater amount of work done.

5. What are some real-life applications of work done by a spring?

Work done by a spring has many practical applications, such as in the suspension systems of vehicles, where springs are used to absorb and dissipate energy from bumps and impacts on the road. Springs are also used in various types of machinery, such as clocks, to store and release energy. Additionally, the concept of work done by a spring is essential in understanding the behavior of elastic materials and structures.

Similar threads

  • Introductory Physics Homework Help
Replies
17
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
461
  • Introductory Physics Homework Help
Replies
12
Views
728
  • Introductory Physics Homework Help
Replies
9
Views
902
  • Introductory Physics Homework Help
Replies
24
Views
1K
  • Introductory Physics Homework Help
Replies
7
Views
932
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
898
  • Introductory Physics Homework Help
Replies
10
Views
950
  • Introductory Physics Homework Help
Replies
3
Views
1K
Back
Top