Potential Energy. Uo = mgh. Apply the Work-Kinetic Energy Theorem: F = - mg x = - (h - ho) ΔK = W = Fx ½ mv 2 - ½ mvo 2 = (-mg ) [- (ho - h)]

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1 Physics 17 Part F Potential Energy U = mgh Apply the Work-Kinetic Energy Theorem: F = - mg x = - (h - ho) ΔK = W = Fx ½ mv 2 - ½ mvo 2 = (-mg ) [- (ho - h)] Re-written: ½ mv 2 + mgh = ½ mvo 2 + mgho Ko = ½ mvo 2 Ko = ½ mvo 2 Uo = mgh U = mgho ( initial kinetic energy) (final kinetic energy) (initial gravitational potential energy) (final gravitational potential energy) Eo = initial total energy E = final total energy E = Eo 1

2 Example A: A falling object s kinetic energy increases by 900 J, i.e., ΔK = 900 J. What is its change in potential energy? U - Uo = Ko - K U - Uo = -(K - Ko) ΔU = - ΔK = -900 J The gain of one of energies equals the losss of the other one: Kinetic energy increases by 900 K; the potential energy decreases by 900 J. The Law of Conservation of Energy Recall from above: ½ mv 2 + mgh = ½ mvo 2 + mgho This is the Law of Conservation of Energy. The sum of the two kinds of energy--the total energy -- doesn t change. Example B: An object of mass m = 10 kg that is 20 meters above the ground is dropped from a tower. (a) What is the initial total energy of the object, assuming elevations are measured relative to the ground? Eo = Ko + Uo = ½ (10) (9.8) 20 = 1960 J (b) What will be the object s speed when its potential energy is 1200 J? K + U = Eo ½ (10)v = 1960 v = m/s 2

3 The conservation of energy equation, E = Eo, is true not only for objects rising and falling straight upward and downward, but also for objects moving horizontally while moving vertically, as is the case for projectile motion, for example, or carts on roller-coasters, or skiers moving up and down hill-sides. Example A: The ball about to fall into the basket is 2.2 meters above the elevation from which it was launched. What is its speed? Using the initial elevation as the reference elevation, we have ½ mv 2 + mg(2.2) = ½ m(7.0) 2 + mg(0) The m s cancel: v = 2.42 m/s Example B: An object at Point A is initially moving at 4.0 m/s on a frictionless hill. What will be its speed when it reaches point C? ½ mv 2 + mg(25) = ½ m(4.0) 2 + mg(30) (Divide by m.) v = m/s Example C: Solve for v using Point B as reference level. ½ mv 2 + mg(15) = ½ m(4.0) 2 + mg(20) v = m/s 3

4 Non-Conservative Work Work done by air resistance or friction doesn t allow the total energy to be conserved, because some of the kinetic energy is transformed into heat energy which exits the system; we speak of that work as being non-conservative work, and symbolize that work as WNC. E = Eo + WNC Example A: Suppose in the previous example, where the final speed was m/s, the actual speed is 9.24 m/s, instead. What would have been the work done by air resistance? Assume the object s mass is 2.0 kg. WNC = E - Eo = [K + U] - [Ko + Uo] = [½ (2.0)(9.24) (9.8) (25)] - [½ (2.0)(4.0) (9.8)(30)] = J Notice that the non-conservative work done is a negative number, as expected, because the frictional force points in a direction that is opposite to the direction of motion. (See the description of work in Figure 2, page 1, Part E Notes.) Spring Potential Energy k = spring constant Units: N/m The stiffer the spring, the larger the spring constant. U = ½ kx 2 Example B: k = 2000 N/m What is the potential energy stored in this spring when it s compressed by 0.40 m? U = ½ (2000)(-0.40) 2 = 160 J Example C: k = 2000 N/m What is the spring potential energy stored in this spring when it s stretched by 0.40 m? U = ½ (2000)(0.40) 2 = 1600 J 4

5 Spring-Mass Systems A spring-mass system is a spring with an object attached to one end of the spring. Note that the potential energy is symmetrical : it is the same whether the spring is stretched or compressed. Example: An object of mass m = 20.0 kg attached to a spring lying on a frictionless tabletop was set in motion by pulling it to the right and releasing it; the object oscillates back and forth. At a certain moment, the object s kinetic energy is 500 J, while the spring potential energy is 900 J. The motion is entirely horizontal, so if we choose the tabletop as the reference level, the gravitational potential energy is always zero and therefore can be ignored. What will be the speed of the object when the potential energy is 300 J? E = Eo K = K = 1100 J ½ (20.0)v 2 = 1100 v = m/s 5

6 Example: An object is dropped from rest toward a spring resting 2 meters below. By how much will the spring be compressed when the object s speed is 5.0 m/s? Use the top of the compressed spring as the reference elevation. K + U grav + U spring = Ko + Uo grav + Uo spring ½ (4)(5) 2 + (4) g (0) + ½ (800) x 2 = ½ (4) (0) (9.8) (2 + x) + ½ (800) (0) 2 x = 0.32 m If we use the top of the uncompressed spring as the reference level, we get the same answer: ½ (4)(5) 2 + (4) g (-x) + ½ (800) x 2 = ½ (4) (0) (9.8) (2) + ½ (800) (0) 2 Note that the final elevation of the object is a negative number. x = 0.32 m 6