Work-Energy Theorum: Spring potential energy vs Kinetic Energy

Senjai

1. The problem statement, all variables and given/known data

A 1350-kg car rolling on a horizontal surface has a speed v = 40 km/h when it strikes a horizontal coiled spring and is brought to rest in a distance of 2.5 m. What is the spring constant of the spring? Ignore Friction and assume spring is mass-less.

2. Relevant equations
[tex] W = \Delta E[/tex]
[tex] E_{pspring} = \frac{1}{2}(kx^2) [/tex]
[tex] E_k = \frac{1}{2}(mv^2) [/tex]


3. The attempt at a solution

First right off the bat, i converted 40 km/h to its m/s equivalent of aprox. 11.11 m/s

i state the law of conservation of energy: Energy before = Energy after

Therefore:

[tex]
E_k = E_{pspring}
\frac{1}{2}(mv^2) = \frac{1}{2}(kx^2)
[/tex]

then i isolate k

[tex] k = \frac{-mv^2}{x^2} [/tex]

now heres the issue, is x negative? because the displacement is against the direction of motion?
and 2.5m = x, (-2.5)^2 gives me a answer of 4266 Nm
but -(2.5)^2 is entirely different.. This has been a long lasting math issue for me.

And what if x is positive?

i know k MUST be positive right?

rl.bhat

(2.5)^2 is correct. There is no negative energy in the nature.

Delphi51

There is no minus sign in mv^2 = kx^2
or in k = mv^2/x^2. No way you can get k negative!
The minus sign in F = -kx is supposed to help keep track of the fact that the force of the spring is opposite to the direction of stretch but it does seem to have a habit of getting in the way. k is ALWAYS positive.

Senjai

Thanks, the negative sign on mv^2 was an algebra error... Thanks for the clarification guys!

vaibhav1803

Your attempt is correct but you missed somethingthat for a spring if you take natural length as the datum, the force on change in length is given as:
[tex] \vec{F}= -k \vec{x} [/tex]

and hence work done by a spring against external forces
[tex] W_{s}=\int\vec{F}\vec{.dx} [/tex]
over the required limits

in our case the answer is
[tex]W_{s}=-\frac{kx^{2}}{2}[/tex]
as
[tex]W_{s}=\Delta E[/tex]
[tex]\Delta E=-\frac{mv^{2}}{2}[/tex]
the change part was where you lost it all...the KE FELL TO ZERO. HENCE A NEGATIVE CHANGE.