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Understanding Energy: Properties and Transfers in Physics

Explore the concept of energy in physics, including its properties, conservation, and transfers through work and impulses. Learn how energy affects objects and discover the principle of Conservation of Total Energy.

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Understanding Energy: Properties and Transfers in Physics

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  1. What is Energy? ･ It’s a property of matter that can be transmitted to other matter; it can “affect” other objects. ･ It’s property of matter such that the total quantity of this property in the universe is constant. That is, it is a conserved quantity. ･ It’s a property of matter that (unlike momentum) can change form. OSU PH 211, Before Class 21

  2. We transfer momentum to an object by exerting a net force on it during a time interval. That is, we exert an impulse on the object: JB = ∫ Fnet.B(t) dt = DPB We transfer energy to an object by exerting a net force on it while it undergoes a spatial displacement in the direction of that force. That is, we do work on the object: WB = ∫ Fs.net.B(s) ds = DEB OSU PH 211, Before Class 21

  3. And, we know what the result of an impulse looks like: JB = ∫ Fnet.B(t) dt = DPB = D[mBvB] So, what does the result of work look like? WB = ∫ Fs.net.B(s) ds = D[??] Do a little mathematical sleuthing, starting with the definition of force as the time rate of change of momentum.… OSU PH 211, Before Class 21

  4. Notice: Fnet.B = dPB/dt = mB(dvB/dt) = mB(dvB/ds)(ds/dt) = mB(dvB/ds)(vB) So: Fnet.Bds = mBvB dvB ∫ Fnet.Bds = mB∫vB dvB ∫ Fnet.Bds = (1/2)mB(vB.f2– vB.i2) OSU PH 211, Before Class 21

  5. Again, we already know what the result of an impulse looks like: JB = ∫ Fnet.B(t) dt = DPB = D[mBvB] And now we know what the result of work looks like: WB = ∫ Fs.net.B(s) ds = D[(1/2)mBvB2] Could we have surmised this result through direct observation (similar to how we first deduced that momentum was P = mv)? Yes.… OSU PH 211, Before Class 21

  6. Transferring Energy by Doing Work If we push on an object of mass m with a steady net force, Fnet, while that object moves through a displacement, x, in the direction of the force, how will the object’s properties change? By Newton’s Laws, we know it will accelerate (and at a steady rate, since Fnet is steady). And we can calculate the resulting motion, using kinematics: (Try with simple numbers: Let m = 3 kg, Fnet = 15 N, x = 4 m, and suppose the object starts with vi = 2 m/s.) OSU PH 211, Before Class 21

  7. If we do this over and over—with any sort of push through any distance—we notice something peculiar: The quantity (1/2)mv2 seems to be the property that changes when a net force is applied to an object while the object moves through some displacement. Since (1/2)mv2 is associated with motion, we call it kinetic energy (specifically, the kinetic energy of translation, KT). Q: What are the units of kinetic energy? Joules: 1 J = 1 kg·m2/s2 Q: And why do we call Joules “energy”? A: Because these same units turn up in all sorts of other phenomena in physics—other ways or abilities to affect other objects. And apparently all those various forms of energy can interchange—but the total number of Joules in the universe doesn’t change. This is the principle of Conservation of Total Energy. OSU PH 211, Before Class 21

  8. A car travels at some speed, v. If the car’s speed then doubles, by what factor does its kinetic energy increase? • K will also double. • K will triple. • K will quadruple. • It depends on the mass of the car. • It depends on the initial speed of the car. OSU PH 211, Before Class 21

  9. A car travels at some speed, v. If the car’s speed then doubles, by what factor does its kinetic energy increase? • KT will also double. • KT will triple. • KT will quadruple. • It depends on the mass of the car. • It depends on the initial speed of the car. KT varies as the square of the object’s speed. OSU PH 211, Before Class 21

  10. Again, the process of giving an object some energy—by pushing on it with a net force while it moves some displacement—is called work. (Note the units of work, force·distance is indeed energy.) Q: Is energy a vector or a scalar quantity? A: It’s a scalar. Energy has no preferred or defined direction. But notice: Work, the transaction done on an object to transfer energy to/from it,is effective only to the extent that the displacement is in the same direction as a net force: Work = F･x = Fx(cos), where  is the angle between the vectors F and x. Thus a given force F may have only a vector component acting as a net force in the direction of the displacement, x. OSU PH 211, Before Class 21

  11. This “s” is important. This really means: OSU PH 211, Before Class 21

  12. Which force does the most work? 1. The 6 N force. 2. The 8 N force. 3. The 10 N force. 4. They all do the same amount of work. OSU PH 211, Before Class 21

  13. Which force does the most work? 1. The 6 N force. 2. The 8 N force. 3. The 10 N force. 4. They all do the same amount of work. OSU PH 211, Before Class 21

  14. A crane lowers a steel girder at a construction site. The girder moves with constant speed. Consider the work Wg done by gravity and the work WT done by the tension in the cable. Which of the following is correct? • 1. Wg is positive and WT is positive. • 2. Wg is negative and WT is negative. • 3. Wg is positive and WT is negative. • 4. Wg and WT are both zero. • 5. Wg is negative and WT is positive. What is the change in KT for the girder? OSU PH 211, Before Class 21

  15. A crane lowers a steel girder at a construction site. The girder moves with constant speed. Consider the work Wg done by gravity and the work WT done by the tension in the cable. Which of the following is correct? • 1. Wg is positive and WT is positive. • 2. Wg is negative and WT is negative. • 3. Wg is positive and WT is negative. • 4. Wg and WT are both zero. • 5. Wg is negative and WT is positive. What is the change in KT for the girder? Zero, because there’s no change in the girder’s mass or speed. OSU PH 211, Before Class 21

  16. Block A (2 kg), initially at rest, falls vertically onto a table from a height of 3 m. Block B (4 kg), also initially at rest, falls vertically onto the table from a height of 2 m. Comparing the two situations at their moments of impact, which is true? 1.KT.A > KT.B and vA > vB. 2.KT.A = KT.B and vA > vB. 3.KT.A < KT.B but vA > vB. 4.KT.A > KT.B but vA < vB. 5. None of the above. OSU PH 211, Before Class 21

  17. Block A (2 kg), initially at rest, falls vertically onto a table from a height of 3 m. Block B (4 kg), also initially at rest, falls vertically onto the table from a height of 2 m. Comparing the two situations at their moments of impact, which is true? 1.KT.A > KT.B and vA > vB. 2.KT.A = KT.B and vA > vB. 3.KT.A < KT.B but vA > vB. 4.KT.A > KT.B but vA < vB. 5. None of the above. As with momentum, you can’t distinguish which object has the most kinetic energy simply by comparing speeds. But where does the kinetic energy come from here? It has been converted from gravitational potential energy. OSU PH 211, Before Class 21

  18. Gravitational Potential Energy An object can possess energy by virtue of its position, not just its motion. And you can “store” that energy by doing work to change the object’s position rather than its motion. This form of energy is called potential energy (U). A familiar example of U is related to an object’s height above the earth. You lift it against the force of gravity, doing positive work on the object; gravity does an equal amount of negative work on it. So the energy you invested isn’t apparent in the object’s motion—but it’s there. It’s stored as gravitational potential energy (UG). This is indeed an investment—not an expenditure: you get the energy back, in the form of kinetic energy, when the object falls. Compare that to, say, friction, which indeed spends (not invests) energy: you can’t get it back. (So where does it go?) OSU PH 211, Before Class 21

  19. Why is doing work against gravity an “investment, rather than an “expense?” Because we know we get it all back, no matter which path we take. Notice that it doesn’t matter how you get from point si to point sf. Forces such as gravity that do path-independent work are called “Conservative Forces.” OSU PH 211, Before Class 21

  20. Example: A 10 kg book, initially at rest, falls vertically for a distance of 2 m onto a table. ･ What work did you need to do to lift the book from the table to the 2-m height? And when it falls from there back to the table, what kinetic energy does it have at impact? ･ What kinetic energy does it have as it falls past the 1 m height? How about the 50-cm height? And what work would you need to do to lift the book from the table to those heights? An object of mass m at any height h above some reference point has more gravitational potential energy (UG) than at that lower reference point, by an amount equal to mgh (assuming constant g here). Q: So… where does an object have zero gravitational potential energy (UG = 0)? A: Anywhere you want—so long as you then measure h as positive in the direction directly opposite to gravity’s force. OSU PH 211, Before Class 21

  21. Mechanical Energy So now when we quantify the energy an object has, we have two identifiable categories: Kinetic energy (K) and potential energy (U). This is like having “ready cash” (K) and “invested funds” (U) that are completely convertible back to “cash.” The sum of all forms of K and U (like summing “sub-accounts” in a bank balance) gives the Mechanical Energy (Emech) of an object: Emech = KT + UG + … The Overall “Bank Balance” The total energy of an object is: Etotal = Emech + Eother OSU PH 211, Before Class 21

  22. There are two basic ways for an object to gain or lose energy: via work (W) or heat flow (Q). So, to account for changes to an object’s energy, we would simply sum the various “deposits” and “with-drawals” (±heat flows and ±work done) on it by external forces: Etotal = Wext + Qext For now, if we ignore heat flow as an energy source—and assume that all external work done on an object affects only its mechanical energy—we get this: (Emech) = Wext So far, this would be: (KT + UG) = Wext Or, in other words: KT + UG = Wext OSU PH 211, Before Class 21

  23. Again: KT + UG = Wext Notice:If the only work done here is the work done by gravity (i.e. only converting between UG and KT), then there was really no “external” deposit or withdrawal made to the bank account—just “internal” rearranging—for as we’ve seen, when gravity alone acts on an object, whatever UG loses, KT gains: KT = –UG That is: KT + UG = 0 Thus: KT + UG = Wext = 0 Conclusion: Wext is the sum of the work done by all forces except gravity (and except any other form of potential energy). To find the change in the overall bank account balance, we don’t need to sum up the work done by potential energy sources, because that work is already accounted for (“shown on the balance sheet,” so to speak). OSU PH 211, Before Class 21

  24. Corollary conclusion: Whenever only “investments”—sources of potential energy (such as gravity)—do work, that’s just an “internal” rearrangement of accounts; we have a case of conservation of mechanical energy (i.e. just a conversion between various forms of kinetic and potential energy): KT + UG = 0 Written out, this is: (KT.f – KT.i) + (UG.f – UG.i) = 0 Or: KT.f + UG.f = KT.i + UG.i Or: (1/2)mvf2 + mghf = (1/2)mvi2 + mghi And when we do have external sources of work, Wext (i.e. when Emech is not conserved), we have: KT + UG = Wext Written out: (KT.f – KT.i) + (UG.f – UG.i) = Wext Or: KT.f + UG.f = KT.i + UG.i + Wext Or: (1/2)mvf2 + mghf = (1/2)mvi2 + mghi + Wext OSU PH 211, Before Class 21

  25. T Example: You pull this 5.0 kg crate (initially at rest) along a rough floor (k = .80) with a rope (T = 75N) at an angle of 30.0° above the floor. What is the speed of the crate after 1.0 meters of pulling? (See the After Class 21 notes for the full solution comparing the energy approach to using Newton’s Laws and kinematics.) OSU PH 211, Before Class 21

  26. Example: A 1000 kg car is rolling slowly across a level surface at 1.00 m/s, heading toward a group of small children. The doors are locked, so you can’t get inside to use the brakes. Instead, you run in front of the car and push on the hood at an angle 30° below horizontal. How hard must you push to stop the car in a distance of 2.0 m? (See the After Class 21 notes for the full solution using the “energy-banking” method—the Work-Energy Theorem.  OSU PH 211, Before Class 21

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