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Class Notes 4, Phyx 2110 Motion and Potential Energy Graphs I. INTRODUCTION 15 Often one is concerned with the motion of an object that is acted on by a single, conservative force. For example, the object might be a satellite in orbit around the Earth where the single, conservative force is the force of gravity on the satellite. For such a force we can define a potential energy PE, and the body speeds up and slows down in such a way that its total mechanical energy ME is conserved, i.e., is constant. In these notes we discuss how a graph of the potential energy vs position can be used to discern the motion of the body. For concreteness, we consider two different conservative forces: one due to the gravitational pull of the Earth (near its surface) and the other due to an ideal spring. There are two important concepts to keep in mind when using a potential energy graph to understand an object’s motion: (1) The negative of the slope of the graph is equal to the force on the object. That is, the force is the negative of the spatial derivative of the potential-energy function. (2) The difference between the total mechanical energy ME of the object (which is constant) and the potential energy PE equals the kinetic energy KE. That is KE = M E − P E. II. GRAVITY The potential energy associated with a mass m at some vertical height y above the Earth’s surface is given by P Eg = mg(y − y0 ). (1) In this expression, y0 can be any height; for example you can choose y0 = 1 m if you wish. Above that point, the potential energy increases and is positive; below it the potential energy decreases and is negative. Don’t worry about the sign of the potential energy: it’s only changes in potential energy that are interesting – and those can be either positive or negative. Everywhere, the slope of the graph is mg. Thus the gravitational force on the mass is −mg (“−” because the force points down; remember, we defined the positive direction to be up). That is, the gravitational force is given by d Fg = − [mg(y − y0 )] = −mg. dy (2) Figure 1 shows the gravitational potential energy for a 1 kg object near the surface of the Earth. Suppose we place such an ogject 0.75 m above y0 (point A on Fig. 10 5 0 -1.5 -1 -0.5 0 0.5 1 1.5 -5 -10 -15 y-y (m) 0 Figure 1. Potential energy mg(y-y0) of a 1 kg mass vs height y-y0 2) and let go. We know that gravity will accelerate the object downwards with acceleration g. As it does so it will lose potential energy P Eg and gain kinetic energy KE in such a way that its total mechanical energy ME remains constant. This motion can be deduced by looking at the potential-energy graph using the two concepts outlined above. Initially KE = 0 and from the graph we see that P Eg = +7.5 J. Therefore ME = +7.5 J . However at point A there is a force acting because the slope is not zero. The object will thus accelerate in the direction of the force, which is negative (downwards). As long as there are no other forces, the objects’s PE will decrease as it falls and its KE will increase. At point B (y − y0 = +0.25 m), for example, P Eg has dropped to +2.5 J; thus KE has increased to 5.0 J [so that KE + PE = +7.5 J (the total mechanical energy) has remained the same]. If the object can fall below the y = y0 mark (don’t be prejudiced: y0 doesn’t have to be at the surface of the Earth, it can be anywhere), P Eg becomes negative. At C (y − y0 = −0.5 m), for example, P Eg equals −5 J, so the KE must equal 12.5 J (that is, 12.5 J + (− 5 J) = +7.5 J). Until the Earth gets in the way (not pictured) the object continues to lose P Eg and gain KE. III. THE SPRING FORCE The potential energy associated with a mass attached to a spring depends on how much the spring is stretched or compressed. Let the equilibrium length of the spring be L0 , and it’s current length be L. L can be greater (stretch) or less (compression) than L0 . The potential energy associated with a spring that is either stretched or compressed is 2 15 1.2 10 1 ME = 7.5 J 0.8 5 C B 0.6 0 -1.5 -1 -0.5 B 0 A ME = 0.49 J 0.5 1 1.5 0.4 -5 0.2 -10 -0.15 -15 -0.1 C -0.05 0 0 0.05 A 0.1 0.15 x (m) y-y (m) 0 Figure 4. Potential energy ½ kx2, illustrating the change in potential energy as the mass moves from A to B to C. Figure 2. Potential energy of 1 kg mass, illustrating the decrease in potential energy as the mass moves from point A to B to C. 1.2 1 0.8 0.6 0.4 0.2 0 -0.15 -0.1 -0.05 0 0.05 0.1 0.15 x (m) Figure 3. Potential energy 1/2 kx2 vs position x. lower curve: k = 40 N/m; upper curve: k = 200 N/m. P Esp = 1 2 k (L − L0 ) , 2 (3) where k is the spring constant – a quantity that measures the stiffness of the spring. (k is measured in N/m). The value of k depends on what the spring is made of and how loosely or tightly coiled the spring is. We usually replace the difference (L − L0 ) by x, so that P Esp = 1 2 kx . 2 (4) As discussed above, the negative of the slope (derivative) of the potential energy is the force. From this equation for P Es p we thus have Fsp d =− dx 1 2 kx 2 = −kx. (5) Figure 3 shows two potential energy graphs, one for an object attached to a spring with spring constant k = 40 N/m (lower curve) and one for an object attached to a spring with spring constant k = 200 N/m (upper curve). In contrast to gravity, where the force is the same at every position (height), for a spring the force sometimes points to the right (for negative values of x — compressions) and sometimes to the left (for positive values of x — stretches). The magnitude of the force is zero (equilibrium) when x = 0 (no compression or stretch) and grows as the magnitude of x grows. Now, let’s see how motion works for an object attached to a spring. The potential energy for an object attached to a spring is pictured in Fig. 4 for a value of k = 200 N/m. We start by stretching the spring by 0.07 m (point A). If the object is released from rest it has a total mechanical energy of +0.49 J. At point A, the potential energy has a positive slope so the object will experience a force in the negative direction. It accelerates, and as it does so it gains KE and loses P Esp . At point B (x = +0.05 m), the objects’s PE is +0.25 J and its KE must be 0.24 J. As the object passes through x = 0 (the position of no stretch) it experiences instantaneously no force (the PE slope is zero there) but it is moving with a KE of 0.49 J. It continues past x = 0 and begins to compress the spring, slowing down as it does so. Finally, at x = −0.07 m (C) the object has lost all of its KE and P Esp is back to +0.49 J. At that instant the object is at rest, but it feels a force pushing to the right (the PE slope is negative, so the force points in the positive direction), so it begins to accelerate to the right. Eventually the object gets to x = +0.07 m again and the whole process repeats over and over. Thus the object oscillates back and forth, with kinetic and potential energy constantly being exchanged as it oscillates.