The potentialenergy stored in the compressed spring of a dart gun, with a spring
ID: 1751509 • Letter: T
Question
The potentialenergy stored in the compressed spring of a dart gun, with a springconstant of 47.50 N/m, is 0.580 J. Find by how much is the springis compressed.A 0.150 kg dart is fired straight up. Find the verticaldistance the dart travels from its position when the spring iscompressed to its highest position.
The same dart is now fired horizontally from a heightof 3.50 m. The dart remains in contact until the spring reaches itsequilibrium position. Find the horizontal velocity of the dart atthat time.
Find the horizontal distance from theequilibrium position at which the dart hits theground.
Explanation / Answer
This problem is nice because all the units are already in there SIform 1.) Equation: KE = (1/2)*k*x^2 0.580 = (1/2) * 47.5 * x^2 x = 0.156 m 2.) Eqn: PE = mgh 0.580 = (.150)*(9.8)*h h = 0.395 m 3.) find it's velocity using KE = (1/2) * m * v^2 0.580 = (1/2) * (0.150) * v^2 v = 2.78 m/s 4.) Need to find the time it takes to fall vertically D = (1/2) * a * t^2 3.5 = (1/2) * 9.8 * t^2 t = 0.845 s Now use: D= v*t D = 2.78 m/s * 0.845 s = 2.35 m Notes: That's a crappy dart gun- Also check my numbers, Iused the windows calc to do them and they might be off Hope that helps!
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