# With what speed does the projectile leave the barrel

A toy gun uses a spring to project a 4.8 g soft rubber sphere horizontally. The spring constant is 12.0 N/m, the barrel of the gun is 14.3 cm long, and a constant frictional force of 0.028 N exists between barrel and projectile. With what speed does the projectile leave the barrel if the spring was compressed 6.5 cm for this launch?

i was wondering if there is an easier way to solve this equation? i did this the first time and got 3.02 m/s but forgot how i did it.

the second time i did it i used this formula V_f = sqrtKX_i^2 - 2fL/m

V_f= sqrt ((12)(.065)^2 - 2(.028)(.143)/(4.8 x 10^-3)) = 2.98

I was wonder if anyone can tell me is there an easier way to doing this in a simple way. I know there is one which gave me the answer 3.02 i just forgot to write it down and forgot it just now.

Are you sure the answer is 2.98? Can you calrify how you arrive at the above eqn?

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the way i got that is

W_nc = (KE + PE_g + PE_s)_f - (KE + PE_g + PEs) = deltaKE + deltaPE_g + deltaPE_s

since gun barrel is horizontal, the gravitational potential energy is constant
deltaPE_g = mg(y_f - y_i) = 0. the elastic potential energy is
PE_s = 1/2kx^2 x = distance spring compress

deltaPE_s = 1/2k(0 - x_i^2) = -1/2kx_i^2

i got this from the book i read.

oh yea forgot this part Wnc = (fcos180*)L = -fL L= length of barrel

-fL = (1/2mv^2 - 0) + (0) + (-1/2kx_i^2) solve for v^2 then input the numbers given.

Well. ya. I do think that energy conservation is the only one method that will allow you to arrive at this answer. What other methods do you think is applicable in this case? Guess, there is no shortcut for this though..

One alternative that would have come to mind would be to use forces. However, ur force is not a constant force. It changes as ur extension of the spring changes, so basically, forces is being ruled out. Thus, the only way to get around this question is by using energy conservation.

Oh thanks a lot. I guess I will have to remember all this for the test.