Solutions for PHY 101 HW#3

1. (a) The period of motion T is given by 2 times pi times the square root of the ratio m/k. Here m is 0.5 kg and k is 20 N/meter. The period works out to be 0.99 sec, or just about 1 second. The frequency is just a shade over 1 cycle/sec, also called 1 Hz.
 Snapshot 0 1 2 3 4 5 6 7 8 Time(sec) 0 1/8 s 1/4 s 3/8 s 1/2 s 5/8 s 3/4 s 7/8 s 1 s Height 0 max 0 min 0 Velocity max 0 min (max downward) 0 max KE max 0 max 0 max PE 0 max 0 max 0

(c) If the maximum height is 0.1 m, then the maximum elastic potential energy is kx2/2 = 20 N/m * (0.1 m)2/2 = 0.1 Joules. The maximum kinetic energy is thus also 0.1 J, a quarter cycle of oscillation earlier or later. Kinetic energy is mv2/2 = 0.1 Joules. Solving for v, we find the maximum speed is 0.63 meters/sec.

2.

(This is a long problem. It is best to recognize that it should be solved in two separate stages.)

The first stage of this problem is to use the information given about the vertical orientation of the spring, with a goal of determining the value of the spring constant of the spring. When the 0.20 kg mass is hung from the spring, its length changes from its “original length” (otherwise known as its rest length) of 5.0 cm to a stretched length of 6.0 cm. In other words, it is stretched by 6.0 cm – 5.0 cm = 1.0 cm = 1.0 * 10-2 m. The force that produces this stretch is the weight of the 0.20 kg mass; Fgrav = mg =
=
0.20 kg * 9.8 m/s2 = 2.0 N. Since we know both the force on the spring and the stretch produced by the force, we can find the spring constant k from the relation F = kx. We find k = F/x = 2.0 N / 1.0 * 10-2 m = 200 N/m.

The next part of the problem concerns oscillatory motion of the same spring and mass, when oriented on a frictionless horizontal surface. The spring is stretched until its total length is 10.0 cm; this means that it has been stretched 10.0 cm – 5.0 cm = 5.0 cm =
5.0 * 10-2 m. This is the position from which it is released to begin its oscillatory motion.

At this amount of stretch, we can find out the amount of elastic potential energy stored in the spring: Uelast = kx2/2 = 200 N/m * (5.0 * 10-2 m)2 / 2 = 0.25 J.

The total energy of a mass on a spring is equal to E = K + Uelast. Just at the instant when the spring is let go, it has velocity = 0, so its kinetic energy is also equal to zero. Thus, the elastic potential energy at the spring’s initial stretch is also the value of the total energy that it will have throughout its oscillation.

The problem asks for the maximum speed of the mass during its oscillation. We can find the answer to this question by recognizing that every time the mass moves through the position of zero stretch for the spring, then the spring will have zero elastic potential energy. At each such moment, the total energy must be all in the form of kinetic energy. Thus, at those moments, K = 0.25 J.

At no moment during the oscillation will there be any greater kinetic energy, since the kinetic energy can’t exceed the value of the total energy. Thus, this value is the maximum kinetic energy. And, since K = mv2/2, then the maximum value of K corresponds to the maximum value of v. Thus, the maximum speed comes when mv2/2 = 0.25 J. Solving for v, we find v = 1.6 m/s.

3.

This physiologically-inspired problem has a physics foundation. (And as we'll see later in the semester, a lot of physiology has a physics foundation.) The biochemical energy contained in the doughnuts can be converted by our bodies into gravitational potential energy. To know how much biochemical energy is stored in these doughnuts in SI units, you need to know that 1 calorie equals 4.186 Joules so that 1 kilcalorie equals 4,186 Joules. Then one glazed doughnut contains 250 times 4,186 J, or 1,046,500 J of biochemical energy, and one jelly doughnut contains 330 times 4,186 J, or 1,381,380 J. If all this energy of one glazed doughnut is converted into an increase in the gravitational potential energy of Sarah, i.e. Sarah climbing a mountain, the height Sarah could attain (assuming that our bodies are 100 percent efficient at this conversion from biochemical to GPE) is equal to (1,046,500 J)/(55 kg*9.8 m/sec2)=1941.6 m. A mile about 1600 m, so she could climb higher than a mile. Wow! After eating the jelly doughnut, Sarah could potentially climb (1,381,380 J)/(55 kg*9.8 m/sec2)=2562.9 m. Wow! To climb only 30 m, Sarah would only need 30 m/2562.9 m = 0.012 of one jelly doughnut. (Don't be fooled by the plural in "doughnuts"). This is about 1/100 of one jelly doughnut---not very much. If the height values seem a bit high (or if you tried to test these numbers by eating a doughnut and climbing a mountain), we have assumed 100 percent efficiency in converting from biochemical energy GPE. However, a lot of the biochemical energy doesn't actually end up as mechanical energy. There are involuntary processes, etc., that cost energy. So only 10-30% of the biochemical energy can actually be converted into GPE. What happens to the other 70-90% of the energy is a deep one, which we'll address more later in the semester. The short answer is that it ends up in the form of heat; but stay tuned.