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Chapter 7 Internal Energy Work done by a spring ❑ We know that work equals force times displacement Fs = −kx Hooke’s Law for the restoring force of an ideal spring. (It is a conservative force.) s = x f − xi But the force is not constant xi xf Fs ,i = −kxi , Fs , f = −kx f Take the average force Fs,avg x=0 Fs,avg Fs,i + Fs, f = 2 = − 12 k(x f + x i ) Then the work done by the spring is ! W s = Fs,avg cos φ s = Fs,avg cos0 s 2 2 1 1 = − 2 k(x f€+ x i )(x f − x i ) = − 2 k(x f − x i ) 2 2 1 1 W s = 2 kx i − 2 kx f =-ΔU 2 i 2 f Ws = kx − kx = U s ,i − U s , f 1 2 1 2 U elastic = U s = kx 1 2 2 Units of N/m m2 = Nm=J ❑ Total potential energy is U total = U g + U s = mgy + kx 1 2 2 Example Problem A block (m = 1.7 kg) and a spring (k = 310 N/m) are on a frictionless incline (θ = 30°). The spring is compressed by xi = 0.31 m relative to its unstretched position at x = 0 and then released. What is the speed of the block when the spring is still compressed by xf = 0.14 m? x=0 xi xf x=0 θ θ Given: m=1.7 kg, k=310 N/m, θ=30°, xi=0.31 m, xf=0.14 m, frictionless Method: no friction, so we can use conservation of energy Initially 2 E = mv + mgh + kx v i = 0, hi = xi sin θ 1 2 1 2 2 i Ei = mgxi sin θ + kx 1 2 2 Finally h f = x f sin θ , find v f 2 2 1 1 E f = 2 mv f + mgx f sin θ + 2 kx f E f = Ei 2 2 1 1 2 mv f + mgx f sin θ + 2 kx f 2 1 = mgxi sin θ + 2 kxi 2 2 2 1 1 2 mv f = mg ( xi − x f ) sin θ + 2 k ( xi − x f ) k 2 2 v f = 2 g ( xi − x f ) sin θ + ( xi − x f ) m 310 2 2 v f = 2(9.8)(0.31 − 0.14) sin 30 + (0.31 − 0.14 ) 1.7 v f = 1.666 + 13.95 = 3.95 ms ! Interesting to plot the potential energies Energy K ➔ Ug xf E=K+U Utotal Us xi xf xi xf xi Molecular potentials (two atoms) Molecule AB Harmonic oscillator potential A+B 0 re UM 1 = ks (r 2 re ) 2 EM Morse potential Better approximation, but need EM, re, α, where ↵= p ks /2EM But at large r, Morse potential does poorly 0 UM Morse UM = EM [1 exp[ ↵(r C6 ! 6 r re )]] 2 Multidimensional potentials C H Map H O 180 600 500 40 35 30 25 20 15 10 5 0 −5 −10 −15 −20 −25 −30 −35 −40 −45 −50 −55 −60 −65 160 400 300 140 200 100 120 0 100 100 θ2 Interaction Potential (cm−1 ) H2-CO, U(R,r1,r2,θ1,θ2,Φ) 7 8 9 R(a0 ) 10 11 12 360 300 240 180 120 60 0 θ1 (degree) r1,r2,θ2,Φ fixed 80 60 R,r1,r2,Φ fixed 40 20 0 0 50 100 150 200 θ1 250 300 350 Example problem If it takes 4.00 J of work to stretch a Hooke’s law spring 10.0 cm from its unstretched length, determine the extra work required to stretch it an additional 10.0 cm. Thermal energy Modeling a solid as a collection of spring-masses Internal energy due to K of atoms and U of “springs” Conservative and Non-conservative Forces ❑ Conservative Force: a force for which the work it does on an object does not depend on the path. Gravity is an example. ❑ We know we can obtain the work with the xf work integral. W = ∫ Fx dx = Wc xi ❑ If the force is conservative, then W=Wc and this work can be related to the change in xf potential energy W = F dx = −ΔU c ∫ xi x y m mg s A A B C h s s mg B θ mg C W = F cos φ s = mgh h W = (mg sin θ ) s = (mg sin θ ) = mgh sin θ W = mgh - Non-conservative Force - a force for which the work done depends on the path - friction - air resistance If the force is conservative, we can find the potential energy due to the force xf U f ( x) = − ∫ Fx dx + U i ( x) it is xi usually convenient to take Ui(x)=0 Or if we know U(x) and the force is conservative, we can obtain F dU ( x) dU ( x) = − Fx dx ⇒ Fx = − dx The x-component of a conservative force equals the negative derivative of the potential energy with respect to x If both conservative and non-conservative forces act on an object, the work-energy theorem is modified Wtotal = WC + WNC = K f − K i WNC = K f − K i − WC For the case of gravity Wg = WC = mg ( yi − y f ) WNC = K f − K i − mg ( yi − y f ) WNC = K f − K i − mgyi + mgy f WNC = ΔK + ΔU = K f − Ki + U f − U i = (K f + U f ) − (Ki + U i ) WNC = E f − Ei = Wsurr ❑ If no net non-conservative forces WNC = 0 ⇒ E f = Ei = E ❑ Then, conservation of mechanical energy holds ΔK = −ΔU Crate on Incline Revisited FN fk fk FN s h mg x θ FBD s θ θ mg WN = 0 Wg = mg sin θ s = mgh = WC W f = − µ k mg cosθ s = WNC ❑ The crate starts from rest, vi=0 K i = 0, Ei = U i = mgh = Wg WNC = E f − Ei E f = Ei + WNC = mgh − µ K mg cosθ s E f < Ei ❑ Some energy, WNC is loss from the system ❑ In this case it is due to the non-conservative friction force → energy loss in the form of heat ❑ Because of friction, the final speed is only 9.3 m/s as we found earlier ❑ If the incline is frictionless, the final speed would be: E f = Ei , since K i = 0,U f = 0 2 1 2 mv f = mgh = Wg 2Wg 2(7510 J) vf = = = 12.3 ms m 100 kg ❑ Because of the loss of energy, due to friction, the final velocity is reduced. It seems that energy is not conserved Conservation of Energy ❑ There is an overall principle of conservation of energy ❑ Unlike the principle of conservation of mechanical energy, which can be “broken”, this principle can not ❑ It says: ``The total energy of the Universe is, has always been, and always will be constant. Energy can neither be created nor destroyed, only converted from one form to another.’’ ❑ So far, we have only been concerned with mechanical energy ❑ There are other forms of energy: heat, electromagnetic, chemical, nuclear, rest mass (Em=mc2) Ei = E f mech others mech others Ei + Ei = Ef + Ef mech mech others others E f = Ei + Ei − E f mech mech E f = Ei + WNC others others ⇒ WNC = Ei − E f = Q total total ❑ Q (WNC) is the energy lost (or gained) by the mechanical system ❑ The electrical utility industry does not produce U=mgh energy, but merely converts energy Lake electricity Light, heat River Hydro-power plant h K=mv2/2 Example Problem A ball is dropped from rest at the top of a 6.10-m tall building, falls straight downward, collides inelastically with the ground, and bounces back. The ball loses 10.0% of its kinetic energy every time it collides with the ground. How many bounces can the ball make and still reach a window sill that is 2.44 m above the ground? Solution: Method: since the ball bounces on the ground, there is an external force. Therefore, we can not use conservation of linear momentum. An inelastic collision means the total energy is not conserved, but we know by how much it is not conserved. On every bounce 10% of K is lost: Qi = 0.1K i , i = 1,2,3,..., n n is the number of bounces Given: h0 = 6.10 m, hf = 2.44 m Eo = U o = mgho E1 = K1 = Eo o 3 6 Since energy is conserved from point 0 to point 1. However, between point 1 and 2, energy is lost Q2 = 0.1K1 WNC = −Q = E final − Einitial − Q2 = E2 − E1 − 0.1K1 = E2 − E1 − 0.1Eo = E2 − Eo E2 = 0.9 Eo = E3 = E4 1 2 4 5 Total energy after one bounce By the same reasoning E5 = 0.9 E4 = 0.9 E2 = 0.9(0.9 Eo ) 2 E5 = 0.9 Eo Total energy after two bounces The total energy after n bounces is then n log(h f / ho ) E f = 0.9 Eo n = n log(0.9) mgh f = 0.9 mgho log(2.44 / 6.10) n n= h f = 0.9 ho log( 0 . 9 ) n 0.9 = h f / ho n = 8.7 n log(0.9) = log(h f / ho ) Answer is 8 bounces n log(0.9) = log(h f / ho ) Power Average power: Pavg W = Δt or Units of J/s=Watt (W) Measures the rate at which work is done F cos φ s Pavg = = F cos φ v avg Δt € Instantaneous power: W dW P = lim = Δt →0 Δt dt ! ! ! ! F ⋅ ds P= = F ⋅v dt W can also be replaced by the total energy E. So that power would correspond to the rate of energy transfer Example A car accelerates uniformly from rest to 27 m/s in 7.0 s along a level stretch of road. Ignoring friction, determine the average power required to accelerate the car if (a) the weight of the car is 1.2x104 N, and (b) the weight of the car is 1.6x104 N. Solution: Given: vi=0, vf=27 m/s, Δt=7.0 s, (a) mg= 1.2x104 N, (b) mg= 1.6x104 N Method: determine the acceleration Pavg W F cos φ s = = Δt Δt ! s ! F ❑ We don’t know the displacement s ❑ The car’s motor provides the force F to accelerate the car – F and s point in same direction Fs mass Need as and s Pavg = = Δt Δt 2 2 (v − v f i) 2 2 v f = v i + 2ass ⇒ ass = 2 2 2 ( % m (v f − v i ) m % 27 2 − 0 ( Pavg = ' *= ' * = 52m Δt & 2 ) (7.0 s) & 2 ) 4 Pavg Pavg 52mg 52(1.2x10 ) 4 = = = 6.4x10 W g 9.80 4 52(1.6x10 ) 4 = = 8.5x10 W (b) 9.80 Or from work-energy theorem 1 2 2 f 2 f W = ΔK = mv W mv Pavg = = Δt 2Δt Same as on previous slide (a)