#### Transcript SPH4U: Lecture 12 Notes

SPH4U Conservation of Energy 1-D Variable Force Example: Spring For a spring we recall that Fx = -kx. F(x) x1 x2 x relaxed position -kx the mass F = - k x1 F = - k x2 Review: Springs Hooke’s Law: The force exerted by a spring is proportional to the distance the spring is stretched or compressed from its relaxed position. FX = -kx Where x is the displacement from the equilibrium and k is the constant of proportionality. relaxed position FX = 0 x More Spring Review F(x) The work done by the spring Ws during a displacement from x1 to x2 is the area under the F(x) vs x plot between x1 and x2. Ws x1 x2 x2 F ( x)dx x1 x2 x Ws -kx In this example it is a negative number. The spring does negative work on the mass (kx)dx x1 1 kx 2 2 x2 x1 1 Ws k x22 x12 2 Problem: Spring pulls on mass. A spring (constant k) is stretched a distance d, and a mass m is hooked to its end. The mass is released (from rest). What is the speed of the mass when it returns to the relaxed position if it slides without friction? m relaxed position m stretched position (at rest) d m after release v m vr back at relaxed position Solution: Spring pulls on mass. First find the net work done on the mass during the motion from x = d to x = 0 (only due to the spring): 1 1 1 2 2 2 2 2 Ws k x2 x1 k 0 d kd 2 2 2 x2 x1 m stretched position (at rest) d m vr relaxed position Problem: Spring pulls on mass. Now find the change in kinetic energy of the mass: 1 1 mv 22 mv12 2 2 1 1 mv 2r m02 2 2 1 mv 2r 2 ΔK x2 x1 m stretched position (at rest) d m vr relaxed position Solution: Spring pulls on mass. Now use work kinetic-energy theorem: Wnet = WS = K. 1 2 1 kd mv r 2 2 2 k vr d m x2 x1 m stretched position (at rest) d m vr relaxed position Springs : Understanding A spring with spring constant 40 N/m has a relaxed length of 1 m. When the spring is stretched so that it is 1.5 m long, what force is exerted on a block attached to the end of the spring? x=0 k x=1 x=0 x = 1.5 k M M FX = -kx (a) -20 N (b) 60 N (c) -60 N FX = - (40N/m) ( .5m) FX = - 20 N Understanding Forces and Motion A block of mass M = 5.1 kg is supported on a frictionless ramp by a spring having constant k = 125 N/m. When the ramp is horizontal the equilibrium position of the mass is at x = 0. When the angle of the ramp is changed to 30o what is the new equilibrium position of the block x1? (a) x1 = 20cm k (b) x1 = 25cm (c) x1 = 30cm x=0 M q = 30o Solution Choose the x-axis to be along downward direction of ramp. FBD: The total force on the block is zero since it’s at rest. Consider x-direction: Force of gravity on block is Fx,g = Mg sin(q Force of spring on block is Fx,s = -kx1 N y x Fx,g = Mg sinq q q Mg Solution Since the total force in the x-direction must be 0: Mg sinq kx1 0 x1 5.1kg 9.81m s 2 0.5 x1 0.2 m 125 N m y x q Μg sin θ k Work by variable force in 3-D: Nice to know explanation Work dWF of a force F acting through an infinitesimal displacement r is: F 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 Work/Kinetic Energy Theorem for a Variable Force in 3D r2 W F dr 12 mv22 12 mv12 kinetic energy r1 Sum up F.dr along path That’s the work integral That equals change in KE For conservative forces, the work is path independent and depends only on starting point and end point 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) R2 Fg(R1) R1 M m 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 R2 R1 m R2 R1 M 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 This can be written as: F.dr r2 U = U2 - U1 = -W = - r1 r1 U1 U2 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. Floor level of 400 Hazel St Science Office (potential is zero here, for sure!) m y2 y1 Wg = -mg y Conservative Forces: We have seen that the work done by gravity does not depend on the path taken. m R2 1 1 Wg GMm R2 R1 R1 M m h Wg = -mgh Understanding 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) The easiest way to solve this problem is to use the W=K property. 2 2RE (b) (c) 3 2 4 3 RE RE Be careful! Solution Since energy is conserved, K = WG. 1 1 WG =GMm R 2 R1 Do not use mgh formula as this only works when h is very small. 2RE RE RE 1 1 ΔK=c R R 2 1 Where c = GMm is the same for both rocks For the first rock: 1 1 1 1 K1 =c c 2 RE R E 2R E For the second rock: 1 1 2 1 K 2 =c c R 3R 3 RE E E 2 K2 3 4 = K1 1 3 2 Conservative Forces: In general, if the work done does not depend on the path taken (only depends the initial and final distances between objects), the force involved is said to be conservative. 1 1 Wg GMm R2 R1 Gravity is a conservative force: Gravity near the Earth’s surface: A spring produces a conservative force: W g mg y 1 Ws k x22 x12 2 Conservative Forces: A force that offers the opportunity of two-way conversion between kinetic and potential energies is called a conservative force. The work done by a conservative force always has these properties: It can always be expressed as the difference between the initial and final values of potential energy function. It is reversible. It is independent of the path of the body and depends only on the starting and ending points. When the starting and ending points are the same, the total work is zero. Conservative Forces: When the only forces that do work are conservative forces, then the total mechanical energy is E = K + U Conservative forces have the nice property of being able to be defined in terms of a potential energy. The usual definition of potential energy is through the work-energy theorem as for kinetic energy, i.e. W = Ui - Uf. Wtotal U1 U2 Wtotal K2 K1 U1 U 2 K 2 K1 U1 K1 U 2 K 2 NonConservative Forces: Not all forces are conservative. Consider the friction force applied to a crate, the total work done by friction force when sliding the crate up a ramp and back down is not zero. (when the direction of the motion reverses so does the friction force, and the friction does negative work in both directions.) Conservative Forces: We have seen that the work done by a conservative force does not depend on the path taken. W2 W 1 = W2 Therefore the work done by a conservative force in a closed path is 0. W1 W2 WNET = W1 - W2 = W1 - W1 = 0 W1 Potential energy change from one point to another does not depend on path Understanding Conservative Forces The pictures below show force vectors at different points in space for two forces. Which one is conservative ? (a) 1 (b) 2 y (c) both y x (1) x (2) Solution Consider the work done by force when moving along different paths in each case: No work is done when going perpendicular to force. WA = WB (1) WA > WB (2) Solution 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 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. Conservative Forces & Potential Energies (stuff you should know): Force F Fg = -mg Fg = GMm R2 Fs = -kx Work (done by force) W -mg(y2-y1) 1 1 GMm R2 R1 1 k x22 x12 2 Change in P.E U = U2 - U1 P.E. function U mg(y2-y1) 1 1 GMm R2 R1 1 k x22 x12 2 mgy + C GMm C R 1 2 kx C 2 (R is the center-to-center distance, x is the spring stretch) Understanding 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) 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) The spring P.E. is twice as big in (2) ! (2) 1 2 The potential energy stored in (1) is: 2 kd 2 kd 2 The potential energy stored in (2) is: 1 2 k 2d 2kd 2 2 Conservation of Mechanical Energy If only conservative forces are present, the total kinetic plus potential energy of is conserved, i.e. the total “mechanical energy” is conserved (def. of ME). (note: E=Emechanical throughout this discussion) E=K+U E = K + U using K = W = W + U = W + (-W) = 0 using U = -W E = K + U is constant!!! Both K and U can change, but E = K + U remains constant. But we’ll see that if dissipative forces act, then energy can be “lost” to other modes (thermal, sound, etc) changing Emechanical and external forces can change Emechanical 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 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 y=0 h1 h2 v 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 Example: The simple pendulum 1/2mv2 will be So at y = 0 maximum at the bottom of the swing. 1/ mv2 = mgh v2 = 2gh1 2 1 v 2 gh1 y y = h1 y=0 h1 v 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 Example: The simple pendulum 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 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 Example: Airtrack & 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 2mgd v m M 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 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 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 d v2 h Hooke’s Law (review) The magnitude of the force exerted by the spring is directly proportional to the distance the spring has moved from its equilibrium. Fx kx Force is opposite to the direction spring is moved Fx kx This is the Force applied to the spring Example A 0.085 kg mass is hung from a vertical spring that is allowed to stretch slowly from its unstretched equilibrium position until it comes to its new equilibrium position 0.20 m below its initial one. a) Determine the force constant of the spring? b) If the ball is returned to the spring’s initial unstreched equilibrium position and then allowed to fall, what is the Net Force on the mass when it has dropped 0.082 m? c) Determine the acceleration of the mass at position b) Solution F=-kx F y 0 mg kx 0 kx mg k mg x M 0.20m N 4.165 m F=mg a) Determine the force constant of the spring? Therefore k= 4.2 N/m 0.085kg 9.8 N kg Solution F=-kx F=mg F N mg kx N N 0.085kg 9.8 4.165 0.082m kg m 0.4915 N b) If the ball is returned to the spring’s initial unstreched equilibrium position and then allowed to fall, what is the Net Force on the mass when it has dropped 0.082 m? Therefore F= 0.49 N Solution F=-kx F y ma y 0.4915 N ma y F=mg 0.4915 N ay 0.085kg m 5.782 2 s c) Determine the acceleration of the mass at position b) Therefore a= 5.8 m/s2 down Elastic Potential Energy (review) The energy stored in objects that are stretched, compressed, bent, or twisted. 1 2 U s kx 2 Understanding A 0.10 kg mass is hung from a vertical spring (k=9.6 N/m). The mass is held so that the spring is at its unstretched equilibrium position. The mass is then allowed to fall. Neglect the mass of the spring. a) How much elastic potential energy is stored in the spring when the mass has fallen 11 cm? b) What is the speed of the mass when it has fallen 11cm? Solution 1 2 kx 2 1 N 2 9.6 0.11m 2 m 5.8 102 J Us x=0 cm x=11 cm M a) How much elastic potential energy is stored in the spring when the mass has fallen 11 cm? Solution x=0 cm M x=11 cm Ei E f 1 2 1 2 mgh mv f kx 2 2 2mgh kx 2 mv 2f 2mgh kx 2 vf m m N 2 2 0.10kg 9.8 2 0.11m 9.6 0.11m s m 0.10kg m 1.0 s b) What is the speed of the mass when it has fallen 11cm? Flash