My Math Forum Conservation of Energy Problem

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 April 26th, 2014, 07:52 PM #1 Newbie   Joined: Nov 2013 Posts: 17 Thanks: 0 Conservation of Energy Problem A dart gun has a spring with a constant of 74 N/m. An 18 g dart is loaded into the gun, compressing the spring from a resting length of 10.0 cm to a compressed length of 3.5 cm. If the spring transfers 75% of its energy to the dart after the gun is fired, how fast is the dart travelling when it leaves the gun so i know: k=74n/m m=0.018kg problem: i dont know what i have to do with the "x" values because one represents the equilibrium and the other represents the compressed state. ' *please show and explain all steps thanks!*
 April 26th, 2014, 08:43 PM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 6,857 Thanks: 2230 Math Focus: Mainly analysis and algebra The distance that determines the potential energy of the compressed spring is the difference between the compressed length and the equilibrium or rest length.
 April 28th, 2014, 02:16 AM #3 Senior Member   Joined: Apr 2014 From: Glasgow Posts: 2,021 Thanks: 666 Math Focus: Physics, mathematical modelling, numerical and computational solutions $\displaystyle k=74$N/m $\displaystyle m = 18$g = $\displaystyle 0.018$kg $\displaystyle \Delta x$ = 10 - 3.5 = $\displaystyle 6.5$cm = $\displaystyle 0.065$m The line above is what v8archie was referring to. It is nice to get into the habit of using $\displaystyle \Delta x$ instead of $\displaystyle x$ whenever you use Hooke's law or potential energy of a spring because it is the change in length of the spring that matters. Just in case you don't know, the 'Delta' ($\displaystyle \Delta$) is a symbol that just means 'change in'. It is given fully by: $\displaystyle \Delta x$ = new length - old length = $\displaystyle x_0 - x$ The new length is the original length of the spring and the old length is the compressed one, because the dart is already loaded and is being fired by the spring. Conservation of energy: $\displaystyle e \times PE = KE$ where $\displaystyle e$ is the efficiency of the energy conversion process (0.75 in our problem), $\displaystyle PE$ is the potential energy stored in the spring and $\displaystyle KE$ is the kinetic energy. Substituting the formulae for the potential energy in the spring and the kinetic energy of the dart: $\displaystyle e \times \frac{1}{2}k(\Delta x)^2 = \frac{1}{2}mv^2$ Now just rearrange for $\displaystyle v$ and insert the numbers

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