Ball being shot out of a spring gun - Velocity

[mybrohshi5]

Homework Statement

The spring has a force constant of 355N/m. The spring is compressed 5.9 cm and the ball has a mass of 0.029kg. The gun is held horizontal. The barrel of the gun is 5.9 cm long so the ball leaves the barrel at the same point it loses contact with the spring. A resisting force of 6.1N acts on the ball as it moves along the barrel.

Find the speed of the ball as it leaves the barrel.

Homework Equations

Ki + Ui + Wf = Kf + Uf

The Attempt at a Solution

Ki = 0 and Uf = 0 so

1/2 kx² + W = 1/2mv²

1/2 (355N/m)(.059²) - 6.1N(.059m) = 1/2(.029kg)(v²)

V = 4.21 m/s

Thank you for any help :)

 

[mybrohshi5]

Ok well i figured it out but i am not sure why the first way i did it doesn't work. maybe someone would be kind enough to explain it to me :)

So the right way to do it is

F_i = kx
F_i = 355N/m (0.59m) = 20.495

F_avg = 20.495 + 0 / 2
F_avg = 10.4725 N

F_total = 10.47 - 6.1
F_total = 4.37 N

W = Fd
W = 4.37N(.059m)
W = .2578 J

W = K = 1/2mv^2

.2578 = 1/2(.029)v^2

V = 4.22 m/s

 

[PhanthomJay]

The force of friction is opposite to the direction of the motion.

 

[mybrohshi5]

Wow i just realized that i made that stupid mistake :)

it is now edited for anyone looking at this in the future.

Thanks Jay.