i-Club - The Ultimate Subaru Resource

Howzit.

This one is for Rich—

Here is a bit more complete explanation of how I arrived at my response to your query last night: “Can [I] calculate the potential energy of that [popper]?” Using a few assumptions, here’s how:

Assumption #1: The popper is moving just as fast when it hits the table as it was the instant it left the table. For a round trip up and down, v_initial = v_final. This, obviously, disregards air drag and assumes that the popper went straight up and down. This one is pretty trivial and doesn’t affect the results too much since the impulse force that is launching the popper positively accelerates it to its highest velocity at a much faster rate than the negative acceleration it experiences due to the force of gravity.

Assumption #2: This is the bigger one and the key to solving this problem. We are going to assume that ALL of the potential energy, U, stored from flipping the popper is converted entirely to kinetic energy, K. This is a big assumption since not all of the U gets converted to K because conservation of energy dictates that the energy needed to make that “popping” sound had to come from somewhere, but as they say, “close enough for government work.”

There is some other minutiae we’ll toss aside along the way that I will acknowledge.

First step is to determine how fast the thing was going when it hit the table. Using the first assumption, we are taking the first half the trip (up) to be symmetric with the second half of the trip (down). We’ll look at the trip going up which has an initial velocity, v_initial, which we’re trying to determine, a final velocity, v_final, of 0 since it goes up and stops before returning, a maximum height, h, and which took some time, t, to complete. Since we don’t have an accurate time measurement for the total translational period, we are going leave t as an unknown which leaves us with the following kinematic relation to utilize:

(v_final)^2 = (v_initial)^2 + 2ah

Acceleration, a, in this case would the acceleration due to gravity and negative since it is slowing down. We have already taken v_final to be 0 so after some algebraic manipulation, we get:

(v_initial)^2 = 2gh

Now that we have a form for v_initial in terms of known quantities, we can move ahead. Using our second assumption, K = U, we solve for K:

K = 0.5*m*(v_initial)^2

where m is the mass of our popper. Plugging in our relation for v_initial we get:

U = K = mgh

Which, interestingly enough, is the definition of the potential energy, U_g, of an object being some distance, h, above a reference point in a gravitational field. [/bow] Plugging and chugging values for g = 9.8 meters per squared seconds and h = 0.5 meters, we arrive at the relation I gave last night of "the potential energy is 4.9 times whatever the mass of the popper is in kilograms, joules."