A particle moving along the x-axis experiences the force shown in the graph. If the particle has 2.0 J of kinetic energy as it passes x = 0 m, what is its kinetic energy when it reaches x = 4 m? J J J J J 11/16/151Oregon State University PH 211, Class #22

A pendulum swings in a vertical circle. At what point does the tension in the string do the most work? 1.Bottom of the loop 2.Top of the loop 3.The tension does no work 11/16/152Oregon State University PH 211, Class #22

A crane lowers a steel girder at a construction site. The girder moves with constant speed. Consider the work W g done by gravity and the work W T done by the tension in the cable. Which of the following is correct? 1. W g is positive and W T is positive. 2. W g is negative and W T is negative. 3. W g is positive and W T is negative. 4. W g and W T are both zero. 5. W g is negative and W T is positive. What is the change in K T for the girder? 11/16/153Oregon State University PH 211, Class #22

11/16/15Oregon State University PH 211, Class #224 A 2 kg book, initially at rest, falls vertically for a distance of 1 m onto a table. What is its kinetic energy just before impact? J J J J J

11/16/15Oregon State University PH 211, Class #225 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. K A > K B and v A > v B. 2. K A = K B and v A > v B. 3. K A v B. 4. K A > K B but v A < v B. 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?

11/16/15Oregon State University PH 211, Class #226 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 (U G ). 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?)

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. It doesn’t matter how you get from point s i to point s f. Forces such as gravity that do path-independent work are called “Conservative Forces.” 11/16/157Oregon State University PH 211, Class #22

11/16/15Oregon State University PH 211, Class #228 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 (U G ) than at that lower reference point, by an amount equal to mgh (assuming constant g here). Q: So… when/where does an object have zero gravitational potential energy?

11/16/15Oregon State University PH 211, Class #229 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 (E mech ) of an object: E mech = K T + U G + … The Overall “Bank Balance” And so the total energy of an object is: E total = E mech + E other

11/16/15Oregon State University PH 211, Class #2210 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 “withdrawals” (the ±heat flows and the ±work done) on it by external influences:  E total = W ext + Q ext 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:  (E mech ) = W ext So far, this would be:  (K T + U G ) = W ext Or, in other words:  K T +  U G = W ext

11/16/15Oregon State University PH 211, Class #2211  K T +  U G = W ext Notice: If the only work done here is the work done by gravity (i.e. only converting between U G and K T ), 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, whatever U G loses, K T gains:  K T = –  U G That is:  K T +  U G = 0 Thus:  K T +  U G = W ext = 0 Conclusion: W ext 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).

11/16/15Oregon State University PH 211, Class #2212 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):  K T +  U G = 0 Written out, this is: (K T.f – K T.i ) + (U G.f – U G.i ) = 0 Or: K T.f + U G.f = K T.i + U G.i Or: (1/2)mv f 2 + mgh f = (1/2)mv i 2 + mgh i And when we do have external sources of work, W ext (i.e. when E mech is not conserved), we have:  K T +  U G = W ext Written out: (K T.f – K T.i ) + (U G.f – U G.i ) = W ext Or: K T.f + U G.f = K T.i + U G.i + W ext Or: (1/2)mv f 2 + mgh f = (1/2)mv i 2 + mgh i + W ext