Calculating Speed of Ball Leaving Spring Gun Barrel

AI Thread Summary
The discussion focuses on calculating the speed of a ball leaving a spring gun barrel, considering both the spring's force and a resisting frictional force. Initially, the velocity of the ball is calculated as 6.93 m/s without friction, and then adjusted to 4.9 m/s when a constant resisting force of 6.00 N is applied. Participants discuss how to determine the position along the barrel where the ball reaches its maximum speed, noting that this occurs before the end of the barrel due to the influence of friction. The key to finding the maximum speed involves accounting for the energy lost to friction, leading to the equation E_kinetic = E_potential - E_friction. The conversation highlights the importance of accurately incorporating all forces and energy transformations in the calculations.
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Homework Statement


The spring of a spring gun has force constant k = 400 and negligible mass. The spring is compressed 6.00 and a ball with mass 0.0300 is placed in the horizontal barrel against the compressed spring. The spring is then released, and the ball is propelled out the barrel of the gun. The barrel is 6.00 long, so the ball leaves the barrel at the same point that it loses contact with the spring. The gun is held so the barrel is horizontal.

I found the velocity as it leaves the barrel as 6.93 m/s
Then it asks:Calculate the speed of the ball as it leaves the barrel if a constant resisting force of 6.00 N acts on the ball as it moves along the barrel.

I got the 4.9 as the right answer.
Then:For the situation in part B, at what position along the barrel does the ball have the greatest speed? (In this case, the maximum speed does not occur at the end of the barrel.)

And:What is that greatest speed?

Homework Equations



F = -kx
W = 1/2kx2
K = 1/2mv2

The Attempt at a Solution


For finding the first velocity, I used W= 1/2kx2
where x = .06 m, and got it to be .72 J.
Then I set it to 1/2mv2 and got 6.93 m/s, which was right.

For part B, where a 6 N force acts against the bullet I found the work of the force with W = F*D = 6*.06 m = .36
Then I subtracted .36 from .72, which was .36, and set .36 = 1/2mv2, and found the velocity to be 4.9, which was right.

For finding the position along the barrel I tried setting .36 = 1/2kx2, and solving for x, but that didn't work, and without that I can't find the last velocity its asking for. Probably a simple mistake but I can't seem to find where, any ideas?? Thanks for any help.
 
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The ball keeps on accelerating when the force of the spring is greater than the force of friction. After that, the friction takes over and the ball decelerates. So you should solve this equation:

kx = F_{\text{friction}}
 
Thanks, I got the position x where the velocity is greatest, but I still can't get the mac velocity. I tried putting that x value in the spring potential energy equation and setting that equal to the kinetic energy and solving for v, but it said the velocity I got was wrong. This question is bugging me because it should be simple but I seem to keep overlooking something.
 
The mistake is that you didn't include the energy wasted to overcome frictional forces. To obtain the velocity you must consider

E_{\text{kinetic}} = E_{\text{potential}} - E_{\text{friction}}
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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