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Physics 111: Lecture 11 Today’s Agenda Review Work done by variable force in 3-D Newton’s gravitational force Conservative forces & potential energy Conservation of “total mechanical energy” Example: pendulum Non-conservative forces friction General work/energy theorem Example problem Physics 111: Lecture 11, Pg 1 Work by variable force in 3-D: Work dWF of a force F acting F through an infinitesimal displacement r is: r . dW = F r The work of a big displacement through a variable force will be the integral of a set of infinitesimal displacements: WTOT = F r . Physics 111: Lecture 11, Pg 2 Work by variable force in 3-D: Newton’s Gravitational Force Work dWg done on an object by gravity in a displacement dr is given by: . ^ . dWg = Fg dr = (-GMm / R2 r) (dR r ^+ Rd) ^ dWg = (-GMm / R2) dR . ^ . (since r ^ = 0, r ^ = 1) r ^ dR ^ ^ r Rd dr Fg m d R M Physics 111: Lecture 11, Pg 3 Work by variable force in 3-D: Newton’s Gravitational Force Integrate dWg to find the total work done by gravity in a “big” displacement: R2 R2 Wg = dWg = (-GMm / R2) dR = GMm (1/R2 - 1/R1) R1 R1 Fg(R2) m R2 Fg(R1) R1 M Physics 111: Lecture 11, Pg 4 Work by variable force in 3-D: Newton’s Gravitational Force Work done depends only on R1 and R2, not on the path taken. 1 1 Wg GMm m R2 R1 R2 R1 M Physics 111: Lecture 11, Pg 5 Lecture 11, Act 1 Work & Energy A rock is dropped from a distance RE above the surface of the earth, and is observed to have kinetic energy K1 when it hits the ground. An identical rock is dropped from twice the height (2RE) above the earth’s surface and has kinetic energy K2 when it hits. RE is the radius of the earth. What is K2 / K1? (a) 2 2RE 3 (b) 2 RE RE (c) 4 3 Physics 111: Lecture 11, Pg 6 Lecture 11, Act 1 Solution Since energy is conserved, K = WG. 1 1 1 1 WG = GMm R R ΔK = c R 2 1 2 R1 Where c = GMm is the same for both rocks 2RE RE RE Physics 111: Lecture 11, Pg 7 1 1 ΔK = c R Lecture 11, Act 1 2 R1 Solution 1 1 1 1 For the first rock: K1 = c c RE 2RE 2 RE 1 1 2 1 For the second rock: K 2 = c c RE 3RE 3 RE 2 K2 3 4 So: = K1 1 3 2 2RE RE RE Physics 111: Lecture 11, Pg 8 Newton’s Gravitational Force Near the Earth’s Surface: Suppose R1 = RE and R2 = RE + y R2 R1 R y RE Wg GMm GMm E m GM y R1R2 R y R R2 E E E GM but we have learned that 2 g RE So: Wg = -mgy m RE+ y RE M Physics 111: Lecture 11, Pg 9 Conservative Forces: We have seen that the work done by gravity does not depend on the path taken. m 1 1 R2 Wg GMm R2 R1 R1 M m h Wg = -mgh Physics 111: Lecture 11, Pg 10 Conservative Forces: In general, if the work done does not depend on the path taken, the force involved is said to be conservative. 1 1 Gravity is a conservative force: Wg GMm R2 R1 Gravity near the Earth’s surface: Wg mgy k x2 x12 1 A spring produces a conservative force: Ws 2 2 Physics 111: Lecture 11, Pg 11 Conservative Forces: We have seen that the work done by a conservative force does not depend on the path taken. W2 W1 = W2 W1 Therefore the work done in a closed path is 0. W2 WNET = W1 - W2 = 0 W1 The work done can be “reclaimed.” Physics 111: Lecture 11, Pg 12 Lecture 11, Act 2 Conservative Forces The pictures below show force vectors at different points in space for two forces. Which one is conservative ? (a) 1 (b) 2 (c) both y y x (1) x (2) Physics 111: Lecture 11, Pg 13 Lecture 11, Act 2 Solution Consider the work done by force when moving along different paths in each case: WA = WB WA > WB (1) (2) Physics 111: Lecture 11, Pg 14 Lecture 11, Act 2 In fact, you could make money on type (2) if it ever existed: Work done by this force in a “round trip” is > 0! Free kinetic energy!! WNET = 10 J = K W=0 W = 15 J W = -5 J W=0 Physics 111: Lecture 11, Pg 15 Potential Energy For any conservative force F we can define a potential energy function U in the following way: W = F.dr = -U The work done by a conservative force is equal and opposite to the change in the potential energy function. r2 U2 This can be written as: F.dr r2 U = U2 - U1 = -W = - r1 r1 U1 Physics 111: Lecture 11, Pg 16 Gravitational Potential Energy We have seen that the work done by gravity near the Earth’s surface when an object of mass m is lifted a distance y is Wg = -mg y The change in potential energy of this object is therefore: U = -Wg = mg y j m y Wg = -mg y Physics 111: Lecture 11, Pg 17 Gravitational Potential Energy So we see that the change in U near the Earth’s surface is: U = -Wg = mg y = mg(y2 -y1). So U = mg y + U0 where U0 is an arbitrary constant. Having an arbitrary constant U0 is equivalent to saying that we can choose the y location where U = 0 to be anywhere we want to. j m y2 y1 Wg = -mg y Physics 111: Lecture 11, Pg 18 Potential Energy Recap: For any conservative force we can define a potential energy function U such that: F.dr S2 U = U2 - U1 = -W = - S1 The potential energy function U is always defined only up to an additive constant. You can choose the location where U = 0 to be anywhere convenient. Physics 111: Lecture 11, Pg 19 Conservative Forces & Potential Energies (stuff you should know): Force Work Change in P.E P.E. function F W(1-2) U = U2 - U1 U ^ Fg = -mg j -mg(y2-y1) mg(y2-y1) mgy + C GMm ^ 1 1 1 1 GMm Fg = r GMm GMm C R2 R1 2 R R R2 R1 k x2 x12 k x2 x12 1 1 1 2 Fs = -kx 2 2 kx C 2 2 2 Physics 111: Lecture 11, Pg 20 Lecture 11, Act 3 Potential Energy All springs and masses are identical. (Gravity acts down). Which of the systems below has the most potential energy stored in its spring(s), relative to the relaxed position? (a) 1 (b) 2 (c) same (1) (2) Physics 111: Lecture 11, Pg 21 Lecture 11, Act 3 Solution The displacement of (1) from equilibrium will be half of that of (2) (each spring exerts half of the force needed to balance mg) 0 d 2d (1) (2) Physics 111: Lecture 11, Pg 22 Lecture 11, Act 3 Solution 1 The potential energy stored in (1) is 2 kd2 kd2 2 1 k2d 2kd2 2 The potential energy stored in (2) is 2 The spring P.E. is twice as big in (2) ! 0 d 2d (1) (2) Physics 111: Lecture 11, Pg 23 Conservation of Energy If only conservative forces are present, the total kinetic plus potential energy of a system is conserved. E=K+U E = K + U = W + U using K = W = W + (-W) = 0 using U = -W E = K + U is constant!!! Both K and U can change, but E = K + U remains constant. Physics 111: Lecture 11, Pg 24 Example: The simple pendulum Suppose we release a mass m from rest a distance h1 above its lowest possible point. What is the maximum speed of the mass and where does this happen? To what height h2 does it rise on the other side? m h1 h2 v Physics 111: Lecture 11, Pg 25 Example: The simple pendulum Kinetic+potential energy is conserved since gravity is a conservative force (E = K + U is constant) Choose y = 0 at the bottom of the swing, and U = 0 at y = 0 (arbitrary choice) E = 1/2mv2 + mgy y h1 h2 y=0 v Physics 111: Lecture 11, Pg 26 Example: The simple pendulum E = 1/2mv2 + mgy. Initially, y = h1 and v = 0, so E = mgh1. Since E = mgh1 initially, E = mgh1 always since energy is conserved. y y=0 Physics 111: Lecture 11, Pg 27 Example: The simple pendulum 1/2mv2 will be maximum at the bottom of the swing. So at y = 0 1/ mv2 = mgh v2 = 2gh1 2 1 v 2 gh1 y y = h1 h1 y=0 v Physics 111: Lecture 11, Pg 28 Example: The simple pendulum Since E = mgh1 = 1/2mv2 + mgy it is clear that the maximum height on the other side will be at y = h1 = h2 and v = 0. The ball returns to its original height. y y = h1 = h 2 y=0 Physics 111: Lecture 11, Pg 29 Example: The simple pendulum Bowling The ball will oscillate back and forth. The limits on its height and speed are a consequence of the sharing of energy between K and U. E = 1/2mv2 + mgy = K + U = constant. y Physics 111: Lecture 11, Pg 30 Example: The simple pendulum We can also solve this by choosing y = 0 to be at the original position of the mass, and U = 0 at y = 0. E = 1/2mv2 + mgy. y y=0 h1 h2 v Physics 111: Lecture 11, Pg 31 Example: The simple pendulum E = 1/2mv2 + mgy. Initially, y = 0 and v = 0, so E = 0. Since E = 0 initially, E = 0 always since energy is conserved. y y=0 Physics 111: Lecture 11, Pg 32 Example: The simple pendulum 1/2mv2 will be maximum at the bottom of the swing. So at y = -h1 1/ mv2 = mgh v2 = 2gh1 2 1 v 2 gh1 y Same as before! y=0 h1 y = -h1 v Physics 111: Lecture 11, Pg 33 Example: The simple pendulum Galileo’s Pendulum Since 1/2mv2 - mgh = 0 it is clear that the maximum height on the other side will be at y = 0 and v = 0. The ball returns to its original height. y y=0 Same as before! Physics 111: Lecture 11, Pg 34 Example: Airtrack & Glider A glider of mass M is initially at rest on a horizontal frictionless track. A mass m is attached to it with a massless string hung over a massless pulley as shown. What is the speed v of M after m has fallen a distance d ? v M m d v Physics 111: Lecture 11, Pg 35 Example: Airtrack & Glider Glider Kinetic+potential energy is conserved since all forces are conservative. Choose initial configuration to have U=0. K = -U 1 m M v 2 mgd 2 v M m d Physics 111: Lecture 11, Pg 36 Problem: Hotwheel A toy car slides on the frictionless track shown below. It starts at rest, drops a distance d, moves horizontally at speed v1, rises a distance h, and ends up moving horizontally with speed v2. Find v1 and v2. v2 d v1 h Physics 111: Lecture 11, Pg 37 Problem: Hotwheel... K+U energy is conserved, so E = 0 K = - U Moving down a distance d, U = -mgd, K = 1/2mv12 Solving for the speed: v1 2 gd d v1 h Physics 111: Lecture 11, Pg 38 Problem: Hotwheel... At the end, we are a distance d - h below our starting point. U = -mg(d - h), K = 1/2mv22 Solving for the speed: v2 2 g d h d-h v2 d h Physics 111: Lecture 11, Pg 39 Non-conservative Forces: If the work done does not depend on the path taken, the force is said to be conservative. If the work done does depend on the path taken, the force is said to be non-conservative. An example of a non-conservative force is friction. When pushing a box across the floor, the amount of work that is done by friction depends on the path taken. » Work done is proportional to the length of the path! Physics 111: Lecture 11, Pg 40 Non-conservative Forces: Friction Suppose you are pushing a box across a flat floor. The mass of the box is m and the coefficient of kinetic friction is k. The work done in pushing it a distance D is given by: Wf = Ff • D = -kmgD. Ff = -kmg D Physics 111: Lecture 11, Pg 41 Non-conservative Forces: Friction Since the force is constant in magnitude and opposite in direction to the displacement, the work done in pushing the box through an arbitrary path of length L is just Wf = -mgL. Clearly, the work done depends on the path taken. Wpath 2 > Wpath 1 B path 1 path 2 A Physics 111: Lecture 11, Pg 42 Generalized Work/Energy Theorem: Suppose FNET = FC + FNC (sum of conservative and non- conservative forces). The total work done is: WNET = WC + WNC The Work/Kinetic Energy theorem says that: WNET = K. WNET = WC + WNC = K WNC = K - WC But WC = -U So WNC = K + U = E Physics 111: Lecture 11, Pg 43 Generalized Work/Energy Theorem: WNC = K + U = E The change in kinetic+potential energy of a system is equal to the work done on it by non-conservative forces. E=K+U of system not conserved! If all the forces are conservative, we know that K+U energy is conserved: K + U = E = 0 which says that WNC = 0, which makes sense. If some non-conservative force (like friction) does work, K+U energy will not be conserved and WNC = E, which also makes sense. Physics 111: Lecture 11, Pg 44 Problem: Block Sliding with Friction A block slides down a frictionless ramp. Suppose the horizontal (bottom) portion of the track is rough, such that the coefficient of kinetic friction between the block and the track is k. How far, x, does the block go along the bottom portion of the track before stopping? d k x Physics 111: Lecture 11, Pg 45 Problem: Block Sliding with Friction... Using WNC = K + U As before, U = -mgd WNC = work done by friction = -kmgx. K = 0 since the block starts out and ends up at rest. WNC = U -kmgx = -mgd x = d / k d k x Physics 111: Lecture 11, Pg 46 Recap of today’s lecture Work done by variable force in 3-D (Text: 6-1) Newton’s gravitational force (Text: 11-2) Conservative Forces & Potential energy (Text: 6-4) Conservation of “Total Mechanical Energy” (Text: 7-1) Examples: pendulum, airtrack, Hotwheel car Non-conservative forces (Text: 6-4) friction General work/energy theorem (Text: 7-2) Example problem Look at Textbook problems Chapter 7: # 9, 21, 27, 51, 77 Physics 111: Lecture 11, Pg 47

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