How does the total energy of the cart change as it goes down the inclined plane?

Transcription

1 Experiment 6 CONSERVATION OF ENERGY AND THE WORK-ENERGY THEOREM In this experiment you will explore the principle of conservation of mechanical energy. You will see that gravitational energy can be converted into kinetic energy. Recall that the gravitational potential energy of an object with mass m situated at a height h from the floor (taken as the zero of energy) is given by U = mgh where g is the acceleration due to gravity. Also, the kinetic energy of an object with mass m and moving with a velocity v is given by K = ½ mv. Objectives: 1. Describe the principle of the conservation of energy.. Apply the principle to situations involving gravitational potential energy and kinetic energy. 3. Make theoretical predictions of the final velocity of a cart after it rolls down an inclined plane using the principle of the conservation of energy. 4. Verify the work-energy theorem 5. Compare your predictions with experimental results. Hypothesis: How does the gravitational potential energy of a cart at the top of an inclined plane compare to the kinetic energy of the cart at the bottom of the inclined plane? How does the total energy of the cart change as it goes down the inclined plane? How does the work done on an object relate to the change in its kinetic energy? Theory: The mechanical energy E of a system is the sum of its potential energy U and the kinetic energy K of the objects within it. In this experiment we will examine what happens to this mechanical energy when only conservative forces (like the gravitational force) cause energy transfers within the system i.e. when frictional and drag forces do not act on the objects in the system. Also, we shall assume that the system is isolated from its environment i.e. no external force from an object outside the system causes energy changes inside the system. When a conservative force does work W on an object within the system, one of these energies increases exactly as much as the other decreases. In an isolated system where only conservative forces cause energy changes, the kinetic energy and potential energy can change, but their sum, the mechanical energy E of the system cannot change. This result is called the principle of conservation of mechanical energy. We can write this principle in one more form as follows: Emech = K + U =0 The principle of conservation of mechanical energy allows us to solve problems that would be quite difficult to solve using only Newton s laws. When the mechanical energy of a system is conserved, we can relate the sum of the kinetic energy and potential energy at any one instant to that at another instant without considering the intermediate motion and without finding the work done by forces involved. Figure 6-1 below shows a block stationary at the top of a frictionless track at a height h from the bottom (where we assume the gravitational potential energy is zero). The mechanical energy of the block at this point is purely gravitational i.e. E = U = mgh. At the bottom of the track the block has acquired a velocity and its gravitational energy is zero. The mechanical energy of the block now is purely kinetic. i.e. E = K = ½mv. At any instant on the incline, the energy of the cart would be part potential and part kinetic. The total energy of the cart however would be the same in all three positions. 1

2 Figure 6-1 The block slides with no friction Work-Energy Theorem When a body with mass m moves under the influence of a constant external force F parallel to its displacement d we say that the force has done some work. The amount of work done is given by: W = Fd If the force applied is not along the direction of displacement of the object we need to take that component of the force along the direction of dis placement to calculate the work done the applied force. In the Figure 6- below, this force would be Fcosα. Hence the work done in moving the block a distance d is given by: W = Fd cosα Figure 6- A constant force does work while acting on a block through a distance The general expression for the work done by a constant force F that causes a displacement s is given by: W = F. s If the applied force is not constant during the displacement then the work done by the force needs to be calculated as follows: x f W = Fdx x i where F is the applied force along the x-axis and x i and x f are the initial and final positions of the mass. The work-energy theorem states that if a mass m, moves along the x axis under the influence of an external force F that points along that axis, then the work done by this force on the mass m as it moves from an initial point x i to a final point x f along the x axis is equal to the change of the kinetic energy of the mass between the two points. This can be written mathematically as:

3 W = K f K i = K If on a horizontal frictionless track a hanging mass pulls a cart that moves along the track, then the work done on the cart must equal the change in the kinetic energy of the cart. We will verify this theorem in the laboratory. See Figure 6-3 Figure 6-3 The falling weight does work on the system, which increases its kinetic energy Examples 1. You drop a.0 kg textbook from a height of 10 m from the ground to a friend who stands on the ground as shown in the Figure 6-4. (a) If the potential energy is taken as being ze ro at ground level, then what is the potential energy of the book when you release it? (b) What is the kinetic energy of the book just before your friend catches it in her outstretched hands (Hand level), which are 1.5 m above ground level. (c) How fast is the book moving as it is caught? Book 10 m Hand level 1.5 m Figure 6-4 A book at rest falls from a height of 10 m. Note that the drawing is not at scale Solution: (a) U = mgh = = 196 J (b) Potential energy of the book at hand level is mgh = = 9.4 J 3

4 Since the total energy is conserved, then the kinetic energy of the book is = J (c) K = J = ½mv.or v = 1.9 m/s. A frictionless roller coaster car tops the first hill in Figure 6-5 with speed v 0. What is its speed at (a) point A, (b) point B and (c) point C? (d) How high will it go on the last hill, which is too high for it to cross? v o A B h h h/ C Figure 6-5 A frictionless roller coaster Solution: Remember the total mechanical energy of the car is E = U + K = mgh + ½mv 0 and that this energy is conserved (a) Since point A is at the same height as that from witch the car began its fall, the speed at this point is also v o (b) E = mg(h/) + ½mv, or mgh + ½mv o = mg(h/) + ½mv Solving for v we get v = gh+ vo (c) E = ½mv. Or mgh + ½mv o = ½mv Solving for v we get v = gh+ vo (d) At the highest point the roller coaster will come to a halt momentarily. At this height (H) the potential energy will be the same as the kinetic energy it had at point C. i.e. Use the velocity calculated in part (c) to find the kinetic energy at point C 1 m( gh+ vo ) = mgh o v H = h + g 4

5 3. A 5.0 kg block moves in a straight line on a horizontal frictionless surface under the influence of a force that varies with position as shown below. See Figure 6-6. How much work does the force do as the block moves from the origin to 8.0 m? Solution: Since the direction of the applied force is along the displacement of the block the work done W = Fd would just be the area under the curve shown above. We can calculate this area in four parts corresponding to the four sections A,B,C and D as seen in the Figure 6 above. Figure 6-6 A force changes as a function of the position of an object Part A: W = 10 = 0 J Part B: Here we need the slope of this section and the form of the equation. viz. we need to know how the force varies with x (i.e. F(x) ) in this region i.e. < x < 4. F(x)= -5x + 0 (N), m < x < 4 m Hence the work done in this section is 4 W = ( 5x+ 0) dx = 10 J Part C: No work is done here as F = 0. Part D: Here we need the slope of this section and the form of the equation. viz. we need to know how the force varies with x (i.e. F(x) ) in this region i.e. 6 < x < 8. Fx ( ) 5 = x+ 6 6 m < x < 8 m 5

6 Hence the work done in this section is: 8 5 W = ( x+ 6) dx = 5J 6 Hence the total work done is 0 J + 10 J - 5 J = 5 J Materials and equipment Balance Wood block (bumper) Inter-phase Pasco 750 Piece of thread of 1. m of length Track with cart Intelligent pulley with clip Mass holder Force sensor Motion sensor Computerized system and program " Data Studio " System of masses Procedure Conservation of energy 1. Make sure that there are two planes as those shown in Figure 6-7 on your lab bench. The plane on the left is slightly inclined, whereas the one on the right is horizontal. The elevation of the inclined plane is exaggerated in Figure 6-7 to make it noticeable. In the right end is the motion sensor. Verify that the horizontal section is made level. Otherwise, use the leveling screw under the track and turn it until the cart remains at rest at any point on the horizontal section Figure 6-7 A cart slides without friction from the elevated end of an inclined plane. Notice that the cart is represented by a block to simplify its drawing 1. Connect the motion sensor to the interface. Calibrate the sensor, if necessary. The instructions to calibrate it are in the appendix, in the section that expla ins how to use DataStudio 3. Select "Graph" in the "Screens" window and choose to plot position as a function of time. 4. Place the cart on the top end of the incline and release it from rest, and click on the Start button to start recording data 5. Stop the cart before it hits the motion sensor and click the Stop button 6. From the graph, find the slope of the diagonal line. Remember that this corresponds to the speed of the cart on the horizontal section of the track. See Figure 6-9 Work and energy theorem 1. Verify that the equipment shown in Figure 6-8 is on you lab bench, and that the intelligent pulley is set up to measure position and speed. Measure the mass of the cart and that of the force sensor 3. Press the button to tare the force sensor while holding with you hand the mass holder suspended at the end of the thread 4. Select a graph of force versus position and another one of speed versus position 5. Places a mass of 10 g as the hanging mass 6. Click the Start button and release the cart from rest at the far end of the track 6

7 7. Hold the cart before it hits against the intelligent pulley and click the Stop button Figure 6-8 A cart travels on a horizontal plane, without friction, due to a constant force Figure 6-9 Graph of position vs. time of a cart in the conservation of energy experiment. Its speed is the slope of the inclined straight line 7

8 Experiment 6. Laboratory Report Conservation of Energy and the Work-Energy Theorem Section Bench Students Names: Conservation of Energy Mass of the cart (m) = Use g = 9.8 m/s Table 1 Run # Height (m) Potential Energy mgh (J) Velocity (v) from the graph (m/s) Kinetic Energy 1 mv (J) % Error Part II: Work Energy Theorem 8. Connect the smart pulley and the force sensor to the computer and as shown in the Figure Measure the masses of the cart and the force sensor (M). 10. Double click the smart pulley icon and choose the variables position and velocity. (The default is velocity) 11. Tare the force sensor when the hanging mass is zero. 1. Choose to graph the Force versus Position. 13. Choose to graph the Velocity versus Position. 14. Place 10 g as the hanging mass. 15. Start the program when the cart is released from rest at the far end of the track. 16. Stop the program when the cart has reached the stopper. Analysis From the Force versus Position graph choose, two positions and find the area under the curve. This is the work done in moving the mass M from position 1 to position. 8

9 From the Velocity versus Position graph find the velocity (V) of the cart at the two positions chosen above and calculate the kinetic energy (½MV ) corresponding to each position. Calculate the change in the kinetic energy when the cart moves from position 1 to position. How does this value compare to the area under the curve you calculated above. Show all your calculations. Repeat the experiment but with 0 g as the hanging mass. Table Mass (M) Hanging Mass Work Done (J) Change in KE (J) % Difference 10 g 0 g Calculations: Conclusions What do you conclude after doing these experiments? 9

10 Questions Figure 6-10 A box on a smooth horizontal surface 1. A force of 10 N is applied to a box at an angle of 37 degrees with the horizontal. The box moves a distance of 5 m. What is the work done by the 10 N force? Assume no friction. See Figure If an applied force varies with position according to F(x) = x - 1, where x is in m, how much work is done by this force on an object that moves from x = 1 m to x = 4 m? 10

11 Questions on Experiment 6 Conservation of Energy and the Work-Energy Theorem This questionnaire has some typical questions on experiment 6. All students who are taking the laboratory course of University Physics I must be able to correctly answer it before trying to make the experiment. 1. A force of 10 N is applied to a box at an angle of 37 degrees with the horizontal. The box moves a distance of 5 m. What is the work done by the 10 N force? Assume no friction. See Figure 6-11 (a) 40 J (b) 30 J (c) 50 J (d) 37.6 J Figure 6-11 A box on a smooth horizontal surface. If an applied force varies with position according to F(x) = 3x 3-5, where x is in m, how much work is done by this force on an object that moves from x = 4 m to x = 7 m? (a) 83 J (b) 1594 J (c) 640 J (d) 76 J 3. A force F = (6i j) N acts on a particle that undergoes a displacement s = (3i + j) m. What is the work done by the force on the particle? (a) 18 J (b) -J (c) 16 J (d) 0 J 4. A force F = (6i j) N acts on a particle that undergoes a displacement s = (3i + j) m. What is the angle between F and s? (a) 56.3º (b) 1.1º (c) 45.8º (d) 36.8º 5. You drop a.0 kg textbook from a height of 10 m from the ground to a friend who stands on the ground as shown in Figure 6-1. (a) If the potential energy is taken as being zero at ground level, then what is the potential energy of the book when you release it? (a) 98 J (b) 0 J (c) 0 J (d) 196 J 6. You drop a.0 kg textbook from a height of 10 m from the ground to a friend who stands on the ground shown in Figure 6-1. What is the kinetic energy of the book just before your friend catches it in her outstretched hands (Hand level), which are 1.5 m above ground level? 11

12 (a) J (b) 9.4 J (c) 196 J (d) 30 J Book 10 m Hand level 1.5 m Figure 6-1 A book at rest falls from a height of 10 m. Note that the drawing is not at scale 7. You drop a.0 kg textbook from a height of 10 m from the ground to a friend who stands on the ground as shown in Figure 6-1. Assume that the potential energy is taken as being zero at ground level. How fast is the book moving as your friend whose hands are 1.5 m above the ground catches it? (a) 14.0 m/s (b) 5.4 m/s (c) 1.9 m/s (d) 0 m/s 8. A 5.0 kg block moves in a straight line on a horizontal frictionless surface under the influence of a force that varies with position as shown below. How much work does the force do as the block moves from the origin to 8.0 m? (a) 5 J (b) -5 J (c) 10 J (d) 0 J 9. A frictionless roller coaster car tops the first hill in Figure 6-13 with speed v o = 10 m/s. What is its speed at point A? The height h in the figure is 5 m. (a) 5 m/s (b) 0 m/s (c) 10 m/s (d) 1 m/s 1

13 10. A frictionless roller coaster car tops the first hill in Figure 6-13 with speed 10 m/s. How high will it go on the last hill, which is too high for it to cross? The height h in the Figure is 5 m. (a) 10.1 m (b) 5.1 m (c) 8. m (d) 0 m v o A B h h h/ C Figure 6-13 A frictionless roller coaster 13