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arturdo968 · Dec 14, 2008, 08:49 AM

melissa fires a spring gun of spring constant 24 N/m that has been compressed 10 cm. A mass of 30 grams (.03kg) was at the end of the spring gun.

a) how fast does the mass leave the gun?(at this point, the spring is no longer compressed)
b) how high does the mass rise?

the answers should be 2.45 m/s and .4m, but I keep getting way off and I've redone it a hundred times, I don't see what can possibly be wrong. Any help?

I have it set up as:
(.5)(.03)(0)^2 + mg(0)+ (.5)(24)(.01)^2=E
E= .0012 J

then I plug that into
E=(.5)(m(v)^2
.0012=.5(.03)(v^2)
v= .28

and that's not right, can anyone see what I'm doing wrong?

Capuchin · Dec 14, 2008, 09:07 AM

10cm = 0.1m, not 0.01m

arturdo968 · Dec 14, 2008, 09:25 AM

Wow, I feel like an idiot. I think I was thinking it was 1 cm. but when I redo the problem with .1m, it comes out to 2.8, not 2.45

Capuchin · Dec 14, 2008, 09:43 AM

2.8m/s looks right to me.

arturdo968 · Dec 14, 2008, 12:36 PM

Okay, then I'm good! Thank you!

visharad · Dec 15, 2008, 02:46 AM

You have ignored the change in gravitational potential energy.
Change in K.E. = 0.5 * m * v^2 = 0.5 * 0.03 * v^2 = 0.015 v^2
Change in spring P.E. = 0 - 0.5*24*0.1^2 = - 0.12 J
Change in gravitational P.E. = 0.03 * 9.8 * 0.1 = 0.0294 J

Therefore 0.015 v^2 - 0.12 + 0.02943 = 0
0.015 v^2 = 0.12 - 0.02943 = 0.0906
v^2 = 0.0906/0.015 = 6.04
v = sqrt(6.04) = 2.458 m/s

visharad · Dec 15, 2008, 02:51 AM

As the spring is released, the mass moves up. So its gravitational potential energy increases. You need to consider that too. That is what I did in my previous post.