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5.3 The Work-Energy Theorem: Kinetic Energy We’ll start with the form of energy that is most closely related to work – kinetic energy (KE). Consider the figure below, in which an object of mass m has an initial speed v0. A constant force F acts on the object (in the same direction as its velocity), increasing the speed to v. The displacement of the object is x. From the last lecture, we know that the work done on the object is = . Then, from chapter 2, we have 2 = 02 + 2, which implies that the object has acceleration = PC141 Intersession 2013 2 −02 . 2 Day 10 – May 28 – WBL 5.3-5.4 Slide 1 5.3 Work-Kinetic Energy Theorem: Kinetic Energy Now, we apply Newton’s second law: 2 − 02 = = 2 Finally, the work done is = = 2 −02 2 = 1 2 2 − 1 02 . 2 We now define the kinetic energy (K) of the object: 1 = 2 2 By this definition, the work done on the object is simply equal to the change in the object’s kinetic energy: = − 0 = ∆ PC141 Intersession 2013 Day 10 – May 28 – WBL 5.3-5.4 Slide 2 5.3 Work-Kinetic Energy Theorem: Kinetic Energy Kinetic energy can be thought of as energy possessed by mass that has motion. It has SI units of Joules, and can not have a negative value (since both m and v2 are either positive or zero). The last equation on the previous slide represents the workkinetic energy theorem: the net work done on a body by all the forces acting on it is equal to the change in kinetic energy of the body When positive net work is done on an object, its kinetic energy increases (which means its speed increases, assuming that mass is constant). When negative net work is done, the object’s kinetic energy and speed must decrease. If net work is zero, the object’s KE and speed do not change. PC141 Intersession 2013 Day 10 – May 28 – WBL 5.3-5.4 Slide 3 5.3 Work-Kinetic Energy Theorem: Kinetic Energy Think back to Problem #2 from the last lecture. We found that when an object is thrown vertically upward, the force of gravity does negative work on the object on the way up (since force and displacement are in opposite directions) and positive work on the way down (since they are in the same direction). Based on the previous paragraph, we can conclude that the object must slow down (lose KE) on the way up and speed up (gain KE) on the way down, which of course is exactly what happens. Then, recall the example on slide 8 from last lecture…we found that the net work done on the sliding block was zero…but we also knew that the block was sliding at constant velocity. Thus, its KE wasn’t changing, which agrees with the result that no work was done. On the other hand, in problems 3 and 4 from last lecture, the objects in question were accelerating, and we found that positive net work was being done in both cases. PC141 Intersession 2013 Slide 4 Day 10 – May 28 – WBL 5.3-5.4 Problem #1: Footrace A father racing his son has half the kinetic energy of the son, who has half the mass of the father. The father speeds up by 1.0 m/s and then has the same kinetic energy as the son. What are the initial speeds of both the father and the son? Solution: In class PC141 Intersession 2013 Day 10 – May 28 – WBL 5.3-5.4 Slide 5 Problem #2: Accelerating from Rest WBL Ex 5.31 A constant net force of 75 N acts on an object initially at rest as it moves through a parallel distance of 0.60 m. a) What is the final kinetic energy of the object? b) If the object has a mass of 0.20 kg, what is its final speed? Solution: In class PC141 Intersession 2013 Day 10 – May 28 – WBL 5.3-5.4 Slide 6 Problem #3: Out-of-Control Truck WBL Ex 5.35 An out-of-control truck with a mass of 5000 kg is traveling at 35.0 m/s when it starts descending a steep 15° incline with locked brakes. The incline is icy, so the coefficient of friction is only 0.30. Use the work-KE theorem to determine how far the truck will skid before it comes to rest. Solution: In class PC141 Intersession 2013 Day 10 – May 28 – WBL 5.3-5.4 Slide 7 Problem #4: Stopped by a Spring A container of mass m = 0.40 kg slides across a horizontal, frictionless counter with speed v = 0.50 m/s. It runs into and compresses a spring of spring constant k = 750 N/m. When the container is momentarily stopped by the spring, by what distance d is the spring compressed? Solution: In class PC141 Intersession 2013 Day 10 – May 28 – WBL 5.3-5.4 Slide 8 5.4 Potential Energy A second type of energy is potential energy (U), named as such because it represents the potential to do work. This potential arises due to the configuration of a system of objects; it can be thought of as “stored work”. This work is done when the position or configuration is altered. Again, consider Problem #2 from the last lecture. As we discussed earlier in this lecture, the object loses KE on the way up and gains KE on the way down. What we didn’t mention is what happens to this KE. As it turns out, the gravitational force converts KE to PE on the way up and then converts PE to KE on the way down. The “system of objects” in this example is the thrown object and the Earth, which interact via the gravitational force. We’ll return to gravity in a bit… PC141 Intersession 2013 Day 10 – May 28 – WBL 5.3-5.4 Slide 9 5.4 Potential Energy For a spring, recall that the work done in stretching or compressing the spring by a distance from an equilibrium 1 position 0 is = 2 (this is the work done by the applied 2 force in causing the stretch, it is not the work done by the oppositely-directed spring force). This work leads to a change in the spring’s potential energy: 1 2 1 2 = ∆ = − 0 = − 0 2 2 For reasons that will soon become clear, we are permitted to define a reference position or configuration where U = 0. Here, we will define this position to be 0 . This tells us that the potential energy of a spring that has been stretched or compressed by a distance x is = PC141 Intersession 2013 Day 10 – May 28 – WBL 5.3-5.4 Slide 10 5.4 Potential Energy Gravitational Potential Energy Consider the situation shown in the figure. A can of mass m is initially at position y0 and then raised to a position y by an applied force F. We assume that the can has a constant velocity during this trip. Therefore, a = 0 and (by Newton’s 2nd law), F = mg. The work done by the applied force in moving the can through a displacement Δy = y - y0 is = ∆ = − 0 = − 0 But we know that = ∆ = − 0 . Therefore, if we set U = 0 at y0, we have the gravitational potential energy U=mgΔy = mgy if y0 = 0 PC141 Intersession 2013 Day 10 – May 28 – WBL 5.3-5.4 Slide 11 5.4 Potential Energy The Zero Reference Point It is important to note that potential energy can only be described relative to a “reference point” or “reference configuration”, where we define U0 = 0. Just as we are free to choose the origin of our axes, we are also free to choose any reference point that we wish; however, some are more convenient than others. For example, in the gravitational example from the last slide, the tabletop is a convenient choice. For a spring, the equilibrium position (where it is neither stretched nor compressed) is convenient. The reason that we have this freedom is that absolute values of U are of no consequence – only changes in U (ΔU) have any physical meaning. PC141 Intersession 2013 Day 10 – May 28 – WBL 5.3-5.4 Slide 12 5.4 Potential Energy In the figure below, a mass moves from the bottom to the top of a short stairwell. On the left, we’ve defined the reference point to be the middle step (here, U0 = 0, and U1 = -mgy, U2 = mgy). During the move, ΔU = U2 - U1 = 2mgy. On the right, the lower step is the middle step. Then, U0 = 0, and U1 = mgy, U2 = 2mgy). During the move, ΔU = U2 - U0 = 2mgy. As we see, the choice of reference position makes no difference in calculating ΔU. PC141 Intersession 2013 Day 10 – May 28 – WBL 5.3-5.4 Slide 13 5.4 Potential Energy Unlike K, U can have a negative value. There’s nothing inherently strange here, it only means that the reference level was defined at a relatively higher potential. This was the case on the previous slide (on the left stairwell). Path-Independence It is also important to point out that changes in potential energy are path-independent. This means that ΔU is only dependent upon the starting and ending positions, and not upon the precise path taken between them. This facilitates the solution of many problems where the exact path taken between two points is complicated or possibly unknown. PC141 Intersession 2013 Day 10 – May 28 – WBL 5.3-5.4 Slide 14 5.4 Potential Energy Chemical Potential Energy Many students in this class are chemistry, biology, or kinesiology majors, and have undoubtedly encountered the concept of chemical potential energy. As any physicist will tell you, there is no difference between CPE and the PE that we have been describing here. CPE is energy stored in the electromagnetic attractions (bonds) between atoms…that is, it is stored in the “configuration of the system of atoms”. During a chemical reaction, the system of atoms is rearranged in some way, and this can result in an increase or decrease in the CPE – perhaps manifested outwardly as a decrease or increase in the system’s temperature. PC141 Intersession 2013 Day 10 – May 28 – WBL 5.3-5.4 Slide 15 Problem #5: Loop-the-Loop In the figure, a small block of mass m = 0.032 kg can slide along the frictionless loop-the-loop, with radius R = 12 cm. The block is released from rest at point P, which is a height h = 5.0R above the bottom of the loop. a) How much work does the gravitational force do on the block as the block travels from point P to point Q, and from point P to the top of the loop? b) If the gravitational potential energy of the block-earth system is taken to be zero at the bottom of the loop, what is that potential energy when the block is at the point P, the point Q, and the top of the loop? Solution: In class PC141 Intersession 2013 Day 10 – May 28 – WBL 5.3-5.4 Slide 16 Problem #6: Change in Potential Energy WBL LP 5.11 A change in gravitational potential energy… A …is always positive B …depends on the reference point C …depends on the path taken by the object D …depends only on the initial and final positions of the object PC141 Intersession 2013 Day 10 – May 28 – WBL 5.3-5.4 Slide 17 Problem #7: Stacking Books WBL Ex 5.39 Six identical books, 4.0 cm thick and each with a mass of 0.80 kg, lie individually on a flat table. How much work would be needed to stack the books one on top of the other? Solution: In class PC141 Intersession 2013 Day 10 – May 28 – WBL 5.3-5.4 Slide 18