Steven W. answered • 08/28/16
Tutor
4.9
(4,314)
Physics Ph.D., professional, easygoing, 11,000+ hours tutoring physics
Hi Ken!
The way I would attack this is by conservation of mechanical energy (mechanical energy (ME) is the sum of the kinetic energy (KE) plus the potential energy (PE) in the system at any moment). The spring force is conservative, which means that -- at any point where the spring force is the only one doing work, as is the case while firing the ball -- the mechanical energy of the ball-spring system is constant.
Let's look at the mechanical energy (MEo) right at the instant when the spring is unlocked and just begins to fire the ball. At this point, the ball is not yet moving (we are neglecting any mass in the spring for all of this, as the problem seems to want us to assume).
MEo = KEo + PEo
At this instant of beginning, there is no kinetic energy in the system because the ball (the only object with appreciable mass) is not moving (note: KE = (1/2)mv2 for an object of mass m with speed v). But the spring has potential energy, which takes the form PEs = (1/2)kx2, where k is the force constant (or spring constant) of the spring, and x is the amount it is compressed.
So, at the start, we have:
MEo = 0 + (1/2)kx2
Once the spring has fully uncoiled, and the ball is about to leave it, we have a different situation. Now, there is no compression on the spring, so x = 0. Thus, PEs = 0 at this point.
But now the ball has some departing speed, v. So, at this point, we have:
MEf = (1/2)mv2 + 0 (where m is the mass of the ball and v is its speed.
Since mechanical energy is constant through this process, ideally, we can say that
MEf = MEo
(1/2)kx2 = (1/2)mv2
One way of describing this is that all the potential energy at the beginning has been transformed to kinetic energy by the time the ball is released. We can then relate the ball's release speed to other given quantities by solving this equation for v.
v = (k/m)1/2x
This gives the departure velocity in terms of other quantities given in the problem.
The force constant is a measure of how much force it takes to stretch or compress the spring some set length. Stiffer springs take more force to stretch or compress a given length, and thus have higher force constants. The force constant is given in units of force/length... which, as presented in this problem, is pounds per inch (the pounds of force required to stretch or compress the spring by one inch). One spring has k1 = 5 lb/in, and the other has k2 = 10 lb/in.
So, for the first spring mentioned, which I label Spring 1:
k1 = 5 lb/in
x1 = 10 in
And for the second spring, Spring 2:
k2 = 10 lb/in
x2 = 5 in
(and the same alloy, same design argument takes out engineering considerations from the final result).
As is mentioned, there two situations give the same load (since F = kx, in magnitude, for a spring). But if you put them into the expression for v above, you may see that they come out with different values (given the (1/m)1/2 is the same for both cases).
I hope this helps you on your way, but if you want to check an answer, or have more questions, just let me know.
Steven W.
tutor
You did just fine with the calculation!
I would say that whatever that equation for v shows is the factor that determine v. If the mass is the same in each case, then (ideally) the compression times the square root of the force constant (x√k) is the determining factor. I do not think there is a specific name for this quantity, but that is what determines v for a ball of mass m shot off several springs.
That expression would also be useful even if you wanted to shoot several different balls off the same spring, for example.
Just let me know if you have any other questions, and thanks for the mental exercise! :)
Report
08/29/16
Ken H.
Steven:
Is there any way to determine actual units for velocity - like inches per second or something? If I assign mass of 1.6 ounces (one tenth pound) where does time enter the equation?
Thank YOU for the mental exercise. I just turned 60, and I had to google search to re-learn how to apply the fractional exponent.
: )
Ken
Report
08/29/16
Steven W.
tutor
My pleasure! I enjoy working out problems like this.
I am not used working in what are called "US Customary units" (most of the problems we have are metric), so I had to look up some information.
First, a quick differentiation between mass and force, and what the units are. Mass, as you may know, is the amount of matter content an object has. A force is something that accelerates that mass (changes its velocity; from rest to motion, from motion to rest, or from one speed to another, for example). Weight, in particular, is the force of gravity on a certain mass. The force of gravity, Fg, on a mass is given by:
Fg = mg
where g is the acceleration due to gravity. In US units, g is usually taken at the Earth's surface to be g = 32 ft/s2.
The pound (lbF) in the US system is a unit of force, and so is the ounce. The unit of mass in the US system is the slug. Because g is standard at the Earth's surface, there is a direct relationship between an object's weight and its mass, which is why we often talk informally about mass in weight units (or weight in mass units, like someone "weighing" 150 kg, even though that is a unit of mass, not force).
But, to make the units work out directly, you would have to define the mass of your ball in slugs. Using the above expression, you can convert from lbs (and thus ounces) to slugs directly... or you can find conversion sites and tables on the internet or in books.
Once you have your mass in slugs, it looks like -- if you convert all your inches to feet -- and use the expression for v we derived above, then your v should come out in ft/s.
If you leave the lengths in inches, it will mismatch with pounds, and you will get out some other unit for velocity that may not even have a name! :)
Report
08/29/16
Ken H.
You have answered my question - and then some.
Thanks again !
Report
08/29/16
Ken H.
BTW: Is there some means available for me to give you a tip?
Report
08/29/16
Steven W.
tutor
Thanks, Ken! Not necessary, of course, but I think the only way it could be done here is to have yup sign up as a student of mine and then I bill you for a session. That doesn't sound very tippy, though! :D But I'm just happy to help out. Let me know if anything else comes up along this line.
Report
08/29/16
Ken H.
Cool. Thanks again!
Report
08/29/16
Ken H.
08/29/16