5 - WorkPowerEnergy - Conservation & Work-Energy Theorem

Unit 5: Work, Power and Energy Total Mechanical Energy, Law of Conservation of Energy & Work-Energy Theorem Total energy (ET) remains the same throughout (if frictionless) Eg = high Ek = zero Eg = high Ek = low Eg = lower Ek = higher Eg = higher Ek = lower Eg = zero Ek = high 5.7 Total Mechanical Energy (ET):  Energy can be transferred or transformed, never lost → Law of Conservation of Energy ET = Eg + Ek ◼If *If friction negligible friction is not negligible then…. ET = Eg + Ek + EFriction Ex#1 A 56kg diver runs & dives from the edge of a 4.0 m cliff. If her speed is 8.0 m/s when she leaves the cliff, what is her speed as she enters the water? Before Jump Eg m = 56 kg g = 9.8 N/kg At Water Level Ek m = 56 kg v = 8 m/s h=4m ET = Eg + Ek Eg m = 56 kg g = 9.8 N/kg = Ek m = 56 kg v=? h=0m ET = Eg + Ek ET = ET Before Jump At Water Level ET = Eg + Ek ET = Eg + Ek Eg = (56)(9.8)(4) Eg = 2195.2 J 1 2 Ek = (56)(8) 2 Ek = 1792J Eg = (56)(9.8)(0) Eg = 0 J ET = 3987.2 J 3987.2 = 0 + Ek Ek = 3987.2 J 1 2 3987.2 = (56)(v) 2 v = 12m / s Ex#2 A child throws a 0.20 kg rock at a tree. As he throws it, it is moving at 20.0 m/s & is located 1.5 m above the ground. If the rock hits the tree at 10.0 m/s, how high is that point above the ground? Throw Tree Eg Ek Eg Ek m = 0.2 kg m = 0.2 kg m = 0.2 kg m = 0.2 kg g = 9.8 N/kg v = 20 m/s g = 9.8 N/kg v = 10 m/s h = 1.5 m ET = Eg + Ek = h= ? ET = Eg + Ek ET = ET Throw ET = Eg + Ek Eg = (0.2)(9.8)(1.5) Eg = 2.94J 1 2 Ek = (0.2)(20) 2 Ek = 40J ET = 42.94J Tree ET = Eg + Ek 1 2 Ek = (0.2)(10) 2 Ek = 10J 42.94 = Eg + 10 Eg = 32.94J 32.94 = (0.2)(9.8)(?) h = 17m Ex#3 A 2.0 kg ball rolls along a frictionless surface. It passes point C at a speed of 20 m/s. What was the speed of the ball at point A? A C 25 m 20 m Point A Eg m = 2 kg g = 9.8 N/kg Point C Ek m = 2 kg v = ? m/s h = 20 m ET = Eg + Ek Eg m = 2 kg g = 9.8 N/kg = Ek m = 2 kg v = 20m/s h = 25 m ET = Eg + Ek ET = ET Point A Point C ET = Eg + Ek ET = Eg + Ek Eg = (2)(9.8)(20) Eg = (2)(9.8)(25) Eg = 490J 1 2 Ek = (2)(20) 2 Ek = 400J Eg = 392J 890 = 392 + Ek Ek = 498J 1 2 498 = (2)(v) 2 v = 22m / s ET = 890J  5.8 Work-Energy Theorem: ◼ The total work done upon an object, between an initial instant & a final instant, is equal to the change in kinetic energy between these two instances ◼ The total work done by forces on an object produces a change in kinetic energy Wtotal = Ek = Ek final − Ekinitial Ex#4 A force of 16 N is applied to a 400.0 g mass, starting at rest, over a distance of 5.0 m. How long does it take the mass to move 5.0 m? Before Eg Eg = 0 After 5m Ek Eg m = 0.4 kg v = 0 m/s Eg = 0 Ek m = 0.4 kg v=? Recall, work = ΔEk So find work applied to move mass 5.0m W = Fd Wapplied = (16)(5) Wapplied = 80J Work required to set block in motion Wapplied = Eapplied = Ek final − Ekinitial 1 1 2 2 80 = (0.4)(v) − (0.4)(0) 2 2 v = 20m / s v2 = v + 2ad 2 2 1 20 = 0 + 2(a)(5) 2 2 a = 40m / s 2 v2 = v1 + at 20 = 0 + (40)t t = 0.50s Ex#5 Point B is 1.2m above Point A. The average force of friction between the 0.15kg slider & the track is 0.11N. The slider travels 2.3m from A to B. You apply a constant force of 6.6N to get it over the hill. What is the minimum distance that you must push the slider to make it over the hill? Top: Bottom: Eg = (0.15)(9.8)(1.2) ET = 2.017J Eg = 1.764J Ek = 0 J Recall, energy= work ET = Wapplied = 2.017J BUT….friction… EFric = WFric = Fd WFric = (0.11)(2.3) EFric = WFric = 0.253J ET = 2.017J 2.017 = (6.6)(d ) d = 0.31m Quantum: HW: pg 346 #4, 5 & 6 pg 353 #1, 2 & 4