Spring Stretch in Car Towing: How Far Does the Spring Extend?

AI Thread Summary
The discussion focuses on calculating the stretch of a spring in a towing scenario involving a car and a 92 kg trailer, with an acceleration of 0.3 m/s² and a spring constant of 2300 N/m. The calculation uses the formula Fs = kx, leading to the equation (92 kg)(0.3 m/s²) = 2300x, resulting in a stretch of x = 0.012 m. Participants confirm that the approach is valid but note it relies on assumptions about unit consistency and constant acceleration. Additionally, they highlight that real-world conditions may cause variations in spring length. The analysis emphasizes the importance of these assumptions in physics problems.
zachattackback
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Homework Statement


A car towing a 92 kg trailor with a spring between the car and the trailor. the accelertion of the car is .3. the spring constant is 2300
how far is the spring stretched


Homework Equations


Fs=kx


The Attempt at a Solution



Fs=kx
ma=kx
(92)(.3)=2300x
x=.012m

i am not sure if that was legal
 
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zachattackback said:

Homework Statement


A car towing a 92 kg trailor with a spring between the car and the trailor. the accelertion of the car is .3. the spring constant is 2300
how far is the spring stretched


Homework Equations


Fs=kx


The Attempt at a Solution



Fs=kx
ma=kx
(92)(.3)=2300x
x=.012m

i am not sure if that was legal

It is legal, but it rests on several assumptions.

1) That the units of a are m/s²

2) That the units of k are N/m

3) That the acceleration of the trailor is constant- the length of the spring is constant. In reality, it would not be if the car started moving with constant acceleration at some point in time.
 
thanks for the help
 

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