Work and Energy. Phys 2211L. Introduction

Transcription

1 Work and Energy Phys 2211L Introduction In this lab you will use the Work-Energy Theorem to predict the motion of a object in three different cases. You will work with your lab partner(s) to develop a expression giving the velocities or forces involved in each case. You will then test your predictions experimentally and analyze your results. The Work-Energy Theorem The Work-Energy theorem is a basic tool in physics. Up to now we have considered the fundamental way to analyze motion: (1) identify the forces acting, (2) apply Newton s Second Law of Motion (F Net = ma ), and (3) apply kinematic relationships so that the acceleration can be converted into information about velocity and position. It turns out that, for any system whatsoever, steps (2) and (3) above can be combined into one step through the Work-Energy Theorem. This is often simpler than going through the vector analysis required in analyzing forces. To be sure, the results give us information about speed (magnitude of velocity) and position rather than velocity and position, but often that is all we need. The theorem can be stated thus: The change in the kinetic energy of a system = the net work done on it: ΔK = W Net

2 Kinetic Energy is energy that an object has by virtue of its mass and its speed. It is defined as: K = 1 2 mv2 where m is the mass and v is the object s speed. Kinetic energy (as well as Work and any form of energy) is measured in joules: 1 joule = 1 newton-meter. For an object at rest (v =0), the kinetic energy is zero. The work-energy theorem tells us that an object s kinetic energy changes as it moves, unless the net work done on it is zero. Roughly speaking, Work is force applied over a distance: Work = Force x Distance. For example, if you exert 5 newtons of force on a box to push it across the floor 2 meters, you have done 10 joules of work on it: W = Fd F More rigorously, we have to consider that the force may not act in the direction that the object moves. If the box is coming toward you and you are trying to stop it, your force would be directed opposite to the motion of the box and the work you would do is then: d v F W = -Fd This means that you do negative work on the box: -10 joules. d

3 In general, we define work as the product of the component of the force in the direction of the motion (Fcosϕ) times the distance the object moves: W = Fdcosϕ F ϕ d More than one force may do work on an object at the same time. The net work done on an object is just the sum of the work done by all forces. What does the work-energy theorem give us? Simply this: If we know the forces acting on an object, then for any distance d the object travels, we can determine its change in speed by calculating the net work the forces do on it. Alternatively, we could determine the forces that are acting by observing the object s change in speed. You will use the theorem in the above form in Experiments 1 and 2. Potential Energy and the Generalized Work-Energy theorem For Experiment 3, it will be convenient to rewrite the theorem to include changes in an object s potential energy. We will call this the generalized work-energy theorem: The change in the mechanical energy of a system = The work done on the system by nonconservative forces.

4 Or: ΔE = W NC The mechanical energy E of an object is the sum of its kinetic energy, K, and its potential energy, U: E = K + U Potential energy is energy an object has by virtue of its position in a force field. The mathematical form of the potential energy depends upon the particular type of force involved. Here, we will only be concerned with gravitational potential energy. Essentially, potential energy is a measure of the kinetic energy the object will develop (in the future!) if the object is released and the force does work on it. It is defined in this (rather backwards sounding) way: the change in potential energy in going from point 1 to point 2 is the negative of the work done by the force as the object moves to point 2. Perhaps a more intuitive way to put it is this: It is the work you would have to do, fighting against the force, to move it to point 2. The prime example is gravitational potential energy. If an object is moved from position (x 1, y 1 ) to position (x 2, y 2 ) (see diagram), the work done by the force of gravity is W = mg(y 2 y 1 ) regardless of the path the object takes, since gravity points downward (opposite to the vertical displacement (y 2 y 1 ) and has no component along the horizontal

5 (x 2, y 2 ) (x 1, y 1 ) leg of the displacement (x 2 x 1 ). So the change in gravitational potential energy in moving to (x 2, y 2 ) is: ΔU = -W = mg y Since only changes in potential energy have physical relevance, we can assign the value U = 0 to any convenient point, for instance to the point (x 1, y 1 ), and measure potential energy relative to that reference position. With this in mind, we can define gravitational potential energy as: U Gravitational = mgh where h is understood to mean the vertical distance above some reference position, where U = 0. Any convenient position can be chosen to be the zero of potential energy. Conservative and Non-conservative forces Not all forces have an associated potential energy. Forces that do are called conservative forces, since under the action of these forces the mechanical energy of an object does not change, regardless of its position or its motion. We say,

6 Energy is conserved. The force of gravity, the spring force, and the electric force are examples of conservative forces. Forces that do not conserve energy are called non-conservative forces. These forces always change the total mechanical energy of an object as it moves. Examples are friction and air resistance, which always dissipate the mechanical energy of an object. (The mechanical energy is turned into heat energy.) The Generalized Work-Energy Theorem We can now rewrite the original theorem in this way: First, break the work done on the system into two parts: the work by conservative forces and the work by non-conservative forces: W net = W C + W NC But the work done by conservative forces is the negative of the change in potential energy: W C = - U, as described above. So we can write the workenergy theorem as K = - U + W NC or K + U = W NC E = W NC This result says: Any change in the mechanical energy of a system must be due to work done on it by non-conservative forces. If such forces do no work on the system (W NC = 0), then the mechanical energy does not change ( E = 0), regardless of how the system moves. This is a very important result.

7 Experiment 1: Atwood s Machine We will use the work-energy theorem, W net = ΔK, to analyze the system below: M 1 d M 2 Two masses, M 1 and M 2 are connected by a string and hung from two pulleys. Let M 1 be greater than M 2, so that when the system is released, M 1 will travel a distance d to the floor. Our objective is to predict the velocity of the system when it impacts the floor. Then we will run the experiment and test our prediction. To do this, we will need to decide how we can find the final velocity of the system experimentally; that is, from distance and time measurements. 1. Develop the prediction. Follow the steps below to develop your prediction. Assume the pulleys are massless and frictionless (and so have no inertia and do no work on the system.) But recognize that air resistance will exert a resistive force on each mass.

8 Draw a Free Body Diagram for each mass, showing the forces acting on it and their direction. Use the symbols R 1 and R 2 for the force of air resistance on each mass. Assume M 1 moves downward and M 2 moves upward; R 1 and R 2 will always point opposite to the direction of motion. Referring to the FBD, write out a work-energy equation for each mass. That is, for each mass, 1. Write down algebraically the work each force will do if the mass moves vertically a distance d. Use the definition: Work = Fdcos ϕ, but assign a sensible variable name for each force. (For instance, let T = tension, R 1 = the force of air resistance on mass 1, etc.) Decide whether cos ϕ = 1, 0 or -1 for each force. 2. The algebraic sum of these individual works is the net work, W net done on each mass. This sum is the left side of your work-energy equation. Write it down. 3. Write down algebraically the change in kinetic energy (K = 1/2 Mv 2 ) that the mass M will undergo on the right side of the expression. Use subscripts to distinguish between initial velocity and the final velocity, and between the two masses. This is ΔK. 4. Set the two side equal: W net = ΔK. Simplify (if possible) and eliminate any zero terms. You now have the work-energy theorem expressed for each mass. Combine the two work-energy equations in order to find the total work done on the system, and the change of kinetic energy of the total system. Eliminate any terms that will be zero when you run the system. For convenience, combine the two air resistance forces into one resistive force R = R 1 + R 2. Solve the combined equation for the final speed. This is your theoretical prediction. Have this approved by your lab instructor before proceeding. Also, derive an experimental expression for the final velocity from the distance and the time of travel. 3. The experiment. Set up the system using an upright rod, Atwood pulleys, and mass hangers. Put 500 grams on each hanger. Set up a starting position for the

9 descending mass use a 2-meter stick for this. Make sure the system can move freely that the pulleys are aligned and there are no forces acting not accounted for in your theory. Experiment to find a value for the total resistive force, R: Place 5 grams on the down-going hanger and set the system in motion with a small push. If the system moves at constant speed (a = 0) after the push, then R must have the same magnitude as the weight of that 5 grams. If the system speeds up or comes to rest before the floor, add or subtract mass from the down-going hanger until it moves at constant speed. The weight of the extra mass added call it the offset mass should be equal to R. (We are assuming that R is a constant force, and that it will have the same value whether or not the system is accelerating.) When ready, run the experiment by placing an extra 10, 15, and 20 grams on the descending hanger. (You can leave the offset mass on the downgoing hanger. Its mass should be included in the total mass for M 1 ). For each mass set, do a practice run first, then time each mass set three times and use the average time in your analysis. Set up a table in your lab manual where you can record the mass used, the times, the experimentally determined final velocities, and your predictions for these. Find the percent difference between the experiment and theoretical results. Post-Lab Analysis. Complete this after the next experiment: 1. Use the generalized work-energy theorem, E = W NC, to predict the final velocity of the system. That is, use potential energy to express the work done by gravity. Show that the same expression results. 2. Use Newton s 2 nd Law to find the acceleration of the system. 3. Show that a = v f 2 /2d. Using this, compare the results from Newton s 2 nd Law and from the work-energy theorem.

10 Experiment 2 Here we will use the generalized Work-Energy Theorem, E = W NC to analyze a mass swinging as a pendulum: θ v A mass on a string is released from rest and swings down from some angle θ. We want to find the speed v of the mass m at any point along its path. It will be convenient for us to release the mass at θ = 90 degrees (the pendulum will start horizontal). We will develop a prediction for its speed as a function of the final angle θ. We will test this for a final angle of θ = 0 degrees, and for a final angle of 45 degrees. 1. Develop your prediction. Do this as a group and summarize your results in your lab report: First, draw a diagram like that above showing the path of the pendulum (an arc of a circle). Label the length of the pendulum L. This is the distance from where the string is tied to the center of the mass. Draw

11 the mass at the 90-degree position (the initial position), and again at some arbitrary angle θ (the final position). Also, decide what point on the path will be the zero of potential energy. Label this point U=0. Now draw a FBD for the mass when it is located at the final position. Identify the force(s) acting on the mass and draw them as arrows in the correct direction. Assume air resistance is negligible. Develop the energy theorem for the mass: 1. Write down algebraically an expression for the change in mechanical energy (E = K + U) the mass will undergo when it swings from 90 degrees to the final angle. Express this in terms of v and h the speed of the mass and its height above the zero of potential energy. Use subscripts to distinguish initial and final values for the kinetic energy (K = 1/2 mv 2 ) and the gravitational potential energy (U = mgh). 2. Looking at your FBD, decide which force(s) are conservative and which are nonconservative. Write down an expression for W NC, the work done by nonconservative forces as the pendulum swings. Decide whether this work is positive, negative or zero and give it the correct sign. 3. Set the two expressions from above equal. Simplify and drop any terms that are zero. Note that we do not need to factor in the work done by gravity here, since that is a conservative force. The work that will be done by gravity is expressed by the potential energy U. Now take it one step further and express the initial and final heights in terms of the length L and the angle(s) θ. Look at your diagram and find the relation between the height h of the mass and L and θ. Finally, solve for the final velocity v. 2. Set up a pendulum and use a photogate timer to test your prediction. You will release the bob at θ = 90 degrees the pendulum will be horizontal. Predict and measure the speed it will have at θ =0. Clamp a pendulum clamp to a table stand. Suspend a cylindrical bob from the pendulum clamp (see diagram below).

12 It is best to use a V-shaped string, as this helps prevent the bob from precessing. The length L of the pendulum is measured from the bottom of the pendulum clamp to the middle of the bob. Measure to the nearest millimeter. Use a length of about 40 cm. Use a spirit level to make sure the pendulum clamp is horizontal. Else you will not have accurate values for h. Use a mass under the table stand to level it. Arrange the photogate so that the middle of the bob will track through the photogate beam when the bob reaches θ = 0. Set the photogate to gate mode. This will display the time interval during which the bob is blocking the beam. (You will have to catch the bob before it swings back through the gate a second time. That would overwrite your recorded time.) Use a Vernier caliper to measure the diameter of the bob and record this. The diameter divided by the time measured by the photogate should be close to the instantaneous speed of the bob as it passes through the gate. Run several trials and use the average time to find the bob s speed. Compare this with your prediction by taking a percent error. 3. Set up a second final position at about 45 degrees. Place the photogate so that the bob will track trough the beam. Tilt the photogate so that the bob travels through the beam along a diameter. Measure the actual angle with an

13 inclinometer placed on the tilted photogate. Compare your theory s prediction with experiment. Post Lab Analysis: Now assume air resistance is not negligible. θ φ Suppose you release a pendulum of length L at 90 degrees and it swings up to an angle φ on the other side before swinging back down. In traveling the arc, the pendulum bob traveled a distance s. Show that the average force of air resistance on the bob must have been mglcos(φ)/s