Suppose you use rubber band A to launch a .01kg object straight up in the air by stretching the rubber band .05 meters. How high will it go?

Question

I was doing a lab in my physics class and I was asked this question on my lab form.

Rubber band A refers to some testing that we did with one rubber band and how much it could stretch a spring with a certain amount of weight.

One test is where the rubber band stretched .03 meters. The force on the rubber band was 1.13 Newtons. with a weight of .1166kg.

The next one was where the rubber band stretched 0.04m. It had a force of 1.71 newtons on it. The mass on the rubber band was .1749kg.

In short, I need help with figuring out how high a 0.01kg object would go in the air with the rubber band being stretched 0.05 meters.

If there is an equation on how to do that I would appreciate it if you would put that in your response too.

Answer 1

The basic equation for a spring and its displacement and associated force is F = kx (you'll often see this as F = -kx but that just means the force is in the opposite direction to the displacement). You'll notice this is just like the linear equation you learned in algebra y = mx + b where the k is equivalent to the slope m. That means k is the slope of the force-displacement curve.

To calculate slope, you use m = (y₁ - y₂)/(x₁ - x₂), so likewise, for this spring, the k = (F₂ - F₁)/(x₂ - x₁). Plug in your lab test values to get k. Assuming you hung a mass on a rubber band, then the force should be mass/g where g = 9.807 so your force numbers (in Newtons) should be 1.143 (0.1166/9.807) and 1.715 (0.1749/9.807). Plug in those values into k = (F₂ - F₁)/(x₂ - x₁) = (1.715 - 1.143)/(0.04 - 0.03) = 57.17 N/m

Now, the potential energy (U) equation for a spring is U = 1/2kx² - that tells you how much energy is in the spring if it is pulled back a certain amount. Releasing that spring (rubber band) will convert that potential energy into kinetic energy of the object being launched and that kinetic energy gets then converted into the potential energy of height as the object goes up.

Since you have now calculated your spring constant k, you can calculate the energy in the rubber band if it is pulled back to 0.05 m (assuming the spring constant is linear over that displacement domain). So U = 1/2(57.17)(0.05)² = 0.07147 joules.

Assuming all that energy gets converted to kinetic energy, the .01 kg object will g straight up until all the kinetic energy gets converted back into the potential energy of height. The potential energy of height (P) is P = mgh where m is the mass of the object being launched, g is the acceleration of gravity (9.807) and h is the height the object goes up.

So 0.07149 = mgh

0.07149 = 0.01(9.807)h

h = 0.07147/(.01*9.807) = 0.7288 m