# Work-Energy Theorum: Spring potential energy vs Kinetic Energy

 P: 104 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 $$W = \Delta E$$ $$E_{pspring} = \frac{1}{2}(kx^2)$$ $$E_k = \frac{1}{2}(mv^2)$$ 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: $$E_k = E_{pspring} \frac{1}{2}(mv^2) = \frac{1}{2}(kx^2)$$ then i isolate k $$k = \frac{-mv^2}{x^2}$$ 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?
 P: 69 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.