Launching Spring Lab

For this lab, we calculated the constant k of a spring, and used that constant, in addition to other data such as the spring's mass, to describe the spring's motion as a projectile in several different circumstances, including determining how to launch the spring a given distance as a projectile and how to launch it so to keep it in the air for a given duration of time.

The first step of the lab was to calculate the spring constant, k, for the spring. To do this, we used the formula F = -kx. We attached one end of the spring to a force sensor and the other end to a distance-measuring mechanism, and we graphed force vs. distance over an assortment of different distances. The resulting values of F vs. x produced an approximation of the graph of the equation of F = -kx for our spring; the slope of the best-fit line of this graph approximation should, therefore, be equal to k.

Here is our graph, with accompanying best-fit line:

Force vs. Distance Pulled

The Y-axis represents Force; the X-axis represents distance pulled

Prelab question #1: Should the line used to fit your data pass through the origin? Explain carefully using Hooke's law. What would the implication be if it dod not pass through the origin?

No, it should not pass through the origin because of the way that we collected data. We measured data from a point at an arbitrary distance from the force probe; the change in distance is therefore the same as it would have been, but it is in the opposite direction, and it does not reach zero at the same point. If we had zeroed the distance at the spring's resting distance, as is the intention of the variable x, then the spring would have gone through the origin; solving using Hooke's law, 0 = -k(0). If the corrected graph did not go through the origin, Hooke's law would be invalid. For any line y = ax, the line must go through the origin; a line that does not go through the origin must have the equation y = ax + b.

The next part of the lab was to sketch Potential Energy of the spring vs. the distance it is stretched:

Potential Energy vs. Distance Pulled

Prelab question #2: Explain why potential energy in a spring is given by the formula above. You will need to relate energy to area under a force vs. displacement graph (as we did in class), and apply this to Hooke's law.

This question is best explained by a diagram, so:

Force vs. Displacement of a Spring

Area = (1/2)Fd is the equation that we get from this graph; the area under a Force vs. Displacement graph will have units of Energy, so Potential Energy_spring = (1/2)Fd. Hooke's law states that F = -kx, and d, in this case, is the same as x from Hooke's law, as both represent the stretch length of the spring; substituting in, one gets that PE_spring = (1/2)(-kx)(x), or PE_spring = -(1/2)kx².

Prelab question #3: Relate energy stored in the spring when compressed to kinetic energy obtained by the spring when it is released. Show that V_launch = xÖ(k/m)

Well, if potential energy pre-launch is conserved, it must equal kinetic energy post-launch; therefore, (1/2)mv² = (1/2)kx² v² = x²(k/m) v = xÖ(k/m)

Prelab question #3, part 2: You now know the velocity of the launch. Derive an expression, in terms of just v, and launch angle, for the horizontal range d of a projectile launched from the ground. As we argued in class, the horizontal range d is maximized when the angle is 45 degrees. Show this by plotting your function on your calculator (choose any value for launch velocity). Find the maximum value of your graph, and verify that it corresponds to an angle of 45 degrees. Include a sketch of this graph with your write-up. Now combine the two equations, and solve for the compression of the spring x (substitute to eliminate v. You should find that:

x = Ö((dmg)/(2k*sinq*cosq))

Couldn't this have been spread out over more than just half a question? Like, maybe, over five questions? Anyways, the expression for velocity, launch angle, and range is, as taken from our in-class notes, d = (V*cosq*2V*sinq)/g. Simplifying, substituting, and solving for x, one gets: d = (V*cosq*2V*sinq)/g d = (V(2*sinq*cosq)/g d = xk(2*sinq*cosq)/mg substituting in the v_launch formula derived earlier 1/d = mg/(xk(2*sinq*cosq) x = (mgd)/(2k*sinq*cosq) x = Ö((mgd)/(2k*sinq*cosq))

The next step was to use our value of k, along with these derivations, to predict the actions of a projectile spring. Specifically, we wanted to get a spring into a bucket 3 meters from the ramp. Solving the above equation,
x = Ö((3 * 0.02083 * 9.8)/(2 * 42.919 * sin(45°) * cos(45°)) x = 0.1195 meters x = 11.95 cm
So x is essentially 12cm.

Using this distance of 12cm., we successfully launched the spring into the bucket, although it took us several tries to get the right angle on the bucket. The next step was to get a spring to stay in the air for a specified length of time; specifically 1 second. To do this, we used the following equation:
V_f = V_i + at xÖ(k/m) = -xÖ(k/m) + 1(g) substituting in for V_i; if an object is launched up, V_i = -V_f 45.38x = -45.38x + g 90.76x = 9.8 x = 0.108 meters
So, to launch the spring into the air for one second, we would have to pull it back 10.8 cm down a 90° ramp.

We did this five times, and added up the actual times for each run and took an average:

0.97

1.12

1.00

1.02

+ 0.92

1.01

As you can see, we came within one one-hundredth of a second of the expected time; given that our timing mechanism was probably accurate to within 1/4 of a second, this is not bad at all.

This concludes the required portion of the lab; we did work on one additional, optional lab segment, and we attained plausible results, but we did not have time to test them. This next portion of the lab involved trying to block a spring in midair, at the peak of its flight. We were told to launch a spring 4 meters, off of a 45° ramp, but instead of letting it travel 4 meters, we were to block its path in midair after 2 meters. To determine the x required to travel 4 meters, we simply plugged into the equation from earlier:
x = Ö((mgd)/(2k*sinq*cosq)) x = Ö((0.02083 * 9.8 * 4)/(2 * 42.9 * sin(45°) * cos(45°))) x = 0.138 meters = 13.8 cm

Given this, it is fairly straightforward to calculate the maximum height:
V_f² = V_i² + 2ad 0 = (xÖ(x/m)*sin(45°))² + 2gd -x²k/m * sin²(45°) = 2(-9.8)(d) 19.60 = 19.6d 1.000 = d
No, really, I didn't plan this! We really would have to hold the blocking mechanism exactly one meter above the top of the ramp! This is plausible, though; the height must be somewhere between 2 meters, the maximum height if the spring were to go straight up at its original angle after being launched, and zero meters, since the spring won't fall below its final height before the end of its trip.

This has been a very long, complex lab. To restate what we learned through this lab, first, there were a great number of equations and derivations, some very long and abstract, such as x = Ö((mgd)/(2k*sinq*cosq)), and some simpler, albeit not too much more intuitive, like v = xÖ(k/m). For the actual lab, we first figured out how to launch a spring three meters (stretch it 12 cm., and launch it off of a 45° ramp), and we then figured out how to get it to stay in the air for 1 second (launch it straight up, stretching it 10.8 cm. beforehand). We really encountered amazingly little error during this lab; the error that did occur was probably due to the inherent inaccuracy of many of the measurements (it's genuinely hard, for example, to measure time with a stopwatch with any degree of accuracy; it's not really feasible to get accuracy to beyond roughly 1/4 of a second), as well as the many factors that we neglected to include in our equations, such as air resistance and friction. Other than these, however, there were no real major sources of error; this is reflected in our results, as they were essentially what we expected.