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.

 

[rdl]

I would like to point out that the answer provided in the attempt at a solution may not be entirely accurate. The equation used, 1/2 kx2 + W = 1/2mv2, is valid only for the case where the ball is released from the spring and there are no other external forces acting on it. However, in this scenario, there is a resisting force of 6.1N acting on the ball as it moves along the barrel.

To accurately calculate the velocity of the ball as it leaves the barrel, we need to take into account this resisting force. This can be done by using the equation F = ma, where F is the net force acting on the ball, m is its mass, and a is its acceleration. In this case, the net force acting on the ball is the difference between the force exerted by the spring (kx) and the resisting force (6.1N). So, we can rewrite the equation as:

kx - 6.1N = ma

Solving for acceleration, we get:

a = (kx - 6.1N)/m

Substituting the values given in the problem, we get:

a = [(355N/m)(0.059m) - 6.1N]/0.029kg = 33.34 m/s2

Now, using the equation v2 = u2 + 2as, where u is the initial velocity (which is 0 in this case), we can calculate the final velocity:

v2 = 0 + 2(33.34 m/s2)(0.059m) = 3.93 m/s

Therefore, the speed of the ball as it leaves the barrel is approximately 3.93 m/s.