|Oct17-10, 08:42 PM||#1|
spring loaded dart gun
1. The problem statement, all variables and given/known data
A spring-loaded dart gun is used to shoot a dart straight up into the air, and the dart reaches a maximum height of 24 meters. The same dart is shot up a second time from the same gun, but this time the spring is compressed only half as far (compared to the first shot). How far up does the dart go this time (neglect friction and assume the spring obeys Hooke's law)?
3. The attempt at a solution
I know I have to use 1/2(k)x^2 , but im not quite sure how to set this problem up.
|Oct17-10, 09:03 PM||#2|
All of your elastic potential energy is going to be transferred to gravitational potential energy; GPE = mgh. Your elastic potential energy, like you mentioned, is going to be 0.5kx^2. Look at the ratios that form; if the spring is compressed only half way, the potential energy is 1/4 of what it was when pulled back fully - which would proportionally change the height.
That makes sense in my head...it's hard to form into words, so I hope that somewhat makes sense!
|Oct17-10, 09:17 PM||#3|
So it would only travel 6 meters?
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