EXAMPLE 8-7 GRADUATION FLING At the end of a graduation ceremony, graduates fling their caps into the air. Suppose a 0.120-kg cap is thrown straight upward with an initial speed of 7.85 m/s, and that frictional forces can be ignored. (a) Use kinematics to find the speed of the cap when it is 1.18 m above the release point. (b) Show that the mechanical energy at the release point is the same as the mechanical energy 1.18 m above the release point. PICTURE THE PROBLEM In our sketch, we choose y = 0 to be at the level where the cap is released with an initial speed of 7.85 m/s, and we choose upward to be the positive direction. In addition, we designate the release 1.18 m - - point as I (Initial) and the point at which y = 1.18 m as f (final). It is the speed at point f that we wish to find. 6 785 m/s REASONING AND STRATEGY a. The cap is in free fall, which justifies the use of constant- O acceleration kinematics. In addition, the cap is thrown straight upward, and hence the motion is one dimensional along the y direction. We want to relate velocity to position; thus we use ed - e2 + 2a Ay (Section 2-5). In this case, - 7.85 m/s, Ay = 1.18 m, and a, = =g. Substituting these values gives ry. b. At the points I and f we calculate the mechanical energy, E - U + K, with U - may for the gravitational potential en- ergy. The kinetic energy, as always, is K - _ me. Known Mass of cap, In - 0.120 kg; initial speed, u, - 7.85 m/s; height above release point, Ay = 1.18m. Unknown (a) Speed of cap, 4 - 7 (b) Mechanical energy at Initial and final points, E - 7, 6 - 7 CONTINUED236 CHAPTER 8 POTENTIAL ENERGY AND CONSERVATION OF ENERGY SOLUTION Part (a) 1. Use kinematics to solve for ey 1 - + 2a, AY 1 - 1Vo, + 20, Ay 2. Substitute e, = 7.85 m/s, Ay = 1.18 m, and a - of to 1 = Vel + 2a Ay find s. Choose the plus sign, since we are interested - V(7.85 m/s) + 2(-9.81 m/s )(1.18 m) = 6.20 m/'s only in the speed: Part (b) 3. Calculate E, . At this point y = 0 and e, = 7.85 m/s: E - U + K - moy, + me - 0 + (0.120 kg)(7.85 m/s)= = 3.70] 4. Calculate E. At this point y = 1.18 m and my = 6.20 m/'s: E - 4 + K - mo + me - (0.120 kg) (9.81 m/s')(1.18 m) + (0.120 kg)(6.20 m/s)# - 1.39J + 2.31J = 3.70] INSIGHT As expected, E, is equal to E. In the remaining Examples in this section we turn this process around; we start with E, - E,, and use this relationship to find a final speed or a final height. As we shall see, this procedure of using energy conservation is a more powerful approach-it actually makes the calculations much simpler. PRACTICE PROBLEM Use energy conservation to find the height at which the speed of the cap is 5.00 m/s. [Answer: 1.87 m] Some related homework problems Problem 27, Problem 29 Finding the Speed at a Given Height An interesting extension of Example 8-7 is shown in FIGURE 8-15. In this case, we are given that the speed of the cap is a at the height y, and we would like to know its speed . when it is at the height y- To find a, we apply energy conservation to the points i and f: 4 + K -4 +K Writing out U and K specifically for these two points yields the following: As before, we cancel m and solve for the unknown speed, of 1 = Vol + 2:(x - x) This result is in agreement with the kinematic equation, of = 6, 2 + 2a, Ay.QUESthi'I 14 (1 point) Retake question A 0.20-kg apple falls from a tree to the ground, 5.22 m below. Ignore air resistance. Take ground level to be y=0. Determine the speed of the apple, in meters per second, when it is 2.46 m above the ground. Study Example 8-7. Use the conservation of mechanical energy {the sum of the gravitational potential energy and kinetic energy is the same throughout the entire motion). What are the kinetic and the potential energies when the apple is just leaving the tree and what are they when the apple is just about to hit the ground