# How do you find acceleration using Hooke's law?

• Sciencelover91
In summary: I just remembered what Newtons stood for so the units cancel out, which leaves me with m/s^2 sorry. Thank you!
Sciencelover91

## Homework Statement

Imagine you were to connect your spring from this experiment to a 1.0 kg lab cart. You pull the car back and stretch the spring 30 cm, then release the car from rest, assuming there is no friction.
A) find the acceleration at the instant when the cart is first released.
B) When the spring's stretch has decreased to 20 cm (thus the cart has moved forward 10 cm), calculate the cart's acceleration at this instant.

## Homework Equations

I have this equation that I solved for from my lab data: Force of spring = (spring constant 7.5N/M) x (the stretch of the spring s). ----> Fos (force on string) = 7.5 N/m (s)
F=ma

## The Attempt at a Solution

a) I know I am given the mass of the cart, initial velocity (0m/s) and the stretch of the string so I tried using f=ma and used Fos = ma.
7.5N/m (.2 m) = (1.0kg) X a
a = 2.25 m/s^2
However I think it's supposed to take into consideration that the initial velocity is zero so I don't know how to apply this without the final velocity.

B) I did force on spring = ma so 7.5(.2m) = 1.0kg x a and got 1.5 m/s^2, I want to know if I did this correctly and if I was supposed to use the fact that it moved 10cm forward.

Your results look fine to me. Remember that it doesn't matter if an object starts out with 0 m/s or 1000000 m/s; any given instantaneous acceleration a = (net F)/m will only tell you how the velocity changes with that time at that instant.

Sciencelover91
Tallus Bryne said:
Your results look fine to me. Remember that it doesn't matter if an object starts out with 0 m/s or 1000000 m/s; any given instantaneous acceleration a = (net F)/m will only tell you how the velocity changes with that time at that instant.
I am a little confused though because from my work, the units do not work out, I do not have seconds in answer and I am not given any information regarding time (seconds) so I assume I would have to incorporate initial/final velocity into my answer. Because otherwise I would be assuming m/s^2 when my work does not show that so I am not sure how to use velocity in here.

Sciencelover91 said:
I am a little confused though because from my work, the units do not work out, I do not have seconds in answer and I am not given any information regarding time (seconds) so I assume I would have to incorporate initial/final velocity into my answer. Because otherwise I would be assuming m/s^2 when my work does not show that so I am not sure how to use velocity in here.
I just remembered what Newtons stood for so the units cancel out, which leaves me with m/s^2 sorry. Thank you!

## 1. How is acceleration related to Hooke's law?

The acceleration of an object is directly proportional to the force applied and inversely proportional to the object's mass. This relationship is described by Newton's second law of motion, which is also the basis of Hooke's law.

## 2. What is Hooke's law and how is it used to find acceleration?

Hooke's law states that the force needed to extend or compress a spring is directly proportional to the distance the spring is stretched or compressed from its equilibrium position. This law can be used to find the acceleration of an object by measuring the force applied and the displacement of the spring.

## 3. Can Hooke's law be applied to objects other than springs?

Yes, Hooke's law can be applied to any elastic material, not just springs. This includes materials like rubber bands, bungee cords, and even human tendons and ligaments.

## 4. Is the acceleration calculated using Hooke's law always constant?

No, the acceleration may vary depending on the force applied and the displacement of the spring. However, if the force and displacement are constant, then the acceleration will also be constant.

## 5. How can Hooke's law be used in real-life applications?

Hooke's law has many practical applications, including in engineering, construction, and medicine. It is used to design and test the strength of materials, and also in devices like shock absorbers, scales, and prosthetic limbs.

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