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Senjai
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[SOLVED] Work-Energy Theorum: Spring potential energy vs Kinetic Energy
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.
[tex] W = \Delta E[/tex]
[tex] E_{pspring} = \frac{1}{2}(kx^2) [/tex]
[tex] E_k = \frac{1}{2}(mv^2) [/tex]
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 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?
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
[tex] W = \Delta E[/tex]
[tex] E_{pspring} = \frac{1}{2}(kx^2) [/tex]
[tex] E_k = \frac{1}{2}(mv^2) [/tex]
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 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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