Work-Energy Theorum: Spring potential energy vs Kinetic Energy

AI Thread Summary
A 1350-kg car traveling at 40 km/h compresses a spring over a distance of 2.5 m, leading to a discussion on the spring constant calculation. The kinetic energy of the car is equated to the potential energy stored in the spring, using the work-energy theorem. The confusion arises regarding the sign of displacement, with clarification that the spring constant (k) must always be positive. The negative sign in the spring force equation indicates the direction of force opposing displacement, but does not affect the positivity of k. Ultimately, the change in kinetic energy is negative as the car comes to rest, confirming the energy transfer to the spring.
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[SOLVED] Work-Energy Theorum: Spring potential energy vs Kinetic Energy

Homework Statement



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.

Homework Equations


W = \Delta E
E_{pspring} = \frac{1}{2}(kx^2)
E_k = \frac{1}{2}(mv^2)


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:

<br /> E_k = E_{pspring}<br /> \frac{1}{2}(mv^2) = \frac{1}{2}(kx^2)<br />

then i isolate k

k = \frac{-mv^2}{x^2}

now here's 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?
 
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(2.5)^2 is correct. There is no negative energy in the nature.
 
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.
 
Thanks, the negative sign on mv^2 was an algebra error... Thanks for the clarification guys!
 
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:
\vec{F}= -k \vec{x}

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

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