ENERGY. Conservative Forces Non-Conservative Forces Conservation of Mechanical Energy Power

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1 ENERGY Conservative Forces Non-Conservative Forces Conservation of Mechanical Energy Power Conservative Forces A force is conservative if the work it does on an object moving between two points is independent on the path of the object between the points. The work done by the force depends only on the initial and the final positions Example The force of gravity W gravity = PE i -PE f 1

2 Non-Conservative Forces A force is non-conservative if the work it does on an object moving between two points depends on the path of the object between the points. Kinetic friction Air drag Examples Conservation of Mechanical Energy E = KE + PE Isolated system when only conservative forces are present, the total mechanical energy is conserved: remains constant KE initial + PE initial = KE final + PE PE final

3 Conservation of Mechanical Energy When non-conservative force are present, the work done by all non-conservative forces equals the change in mechanical energy of the system. W nc W nc nc = (KE( final + PE final ) (KE initial + PE KE final PE final KE initial nc = (KE( final KE initial ) + ( PE final PE KE final KE initial PE final W nc = KE+ PE PE initial ) PE initial ) Conservation of Mechanical Energy Problem 7: Example 1 A child and sled with combined mass of 50.0 kg slide down a frictionless slope. If the sled starts from rest and has a speed of 3.00 m/s at the bottom, what is the height of the hill? 1. Only conservative forces here!. Take the zero potential energy at the base of the hill KE initial + PE initial = KE final + PE 0 J + mgy i = ½ mv f + 0 J v f y = g i = m PE final A 0 y A: initial position KE i = 0 J (v i = 0 m/s) PE i = mgy i B: final position KE i = ½mv f B PE i = 0 J (y f = 0 m) 3

4 Conservation of Mechanical Energy Problem 8: Example A kg bead slides on a curved wire, starting from rest at point A. If the wire is frictionless, find the speed of the bead a) at point B b) at point C 1) Only conservative forces here! ) Take the zero potential energy at point B Conservation of Mechanical Energy A: initial position KE i = 0 J (v A = 0 m/s) PE i = mgy A y A = 5.00 m Example a) B: final position KE f = ½mv B PE f = 0 J (y B = 0 m) KE final + PE final =KE initial + PE initial ½ mv B + mgy B = ½ mv A + mgy A ½ mv B + 0 = 0 + mgy A v B = gy A v B = 9.90 m/s 4

5 Conservation of Mechanical Energy Example b) A: initial position KE i = 0 J (v A = 0 m/s) PE i = mgy A y A = 5.00 m C: final position KE f = ½mv C PE f = = mgy C y C =.00 m KE final + PE final =KE initial + PE initial ½ mv C + mgy C = ½ mv A + mgy A v C = g(y A y c ) v C = 7.67 m/s Conservation of Mechanical Energy Problem 30: Example 3 A bead of mass m = 5.00 kg is released from point A and slides on a frictionless track. Determine a) the bead s speed at points B and C b) the net work done by the force of gravity in moving from A to C 5

6 Conservation of Mechanical Energy A: initial position KE i = 0 J (v A = 0 m/s) PE i = mgy A y A = 5.00 m Example 3 a) C: final position KE f = ½mv C PE f = mgy C y C =.00 m PE final KE initial PE initial KE final + PE final =KE initial + PE initial ½ mv C + mgy C = ½ mv A + mgy A ½ mv C + mgy C = 0 + mgy A v C = g (y A y C ) v C = 7.67 m/s Conservation of Mechanical Energy A: initial position KE i = 0 J (v A = 0 m/s) PE i = mgy A y A = 5.00 m Example 3 a) B: final position KE f = ½mv B PE f = mgy B y B = 3.0 m PE final KE initial PE initial KE final + PE final =KE initial + PE initial ½ mv B + mgy B = ½ mv A + mgy A ½ mv B + mgy B = 0 + mgy A v B = g (y A y B ) v B = 5.94 m/s 6

7 Conservation of Mechanical Energy Example 3 b) The work done by gravity when the bead is moving from point A to point C is: W gravity = PE initial PE final W gravity = PE A PE C W gravity = mgy A mgy C W gravity = mg (y A y C ) W gravity = 147 J Potential Energy Stored in a Spring a) Spring at equilibrium F b F s b) Spring is compressed Force exerted on spring by the block is F b = k x Force exerted by the spring on block is F s = - kx k is the spring constant (Units: N/m) Hooke s Law : F s = - kx (restoring force opposite to displacement) c) Block is released. PE stored in spring transferred to block in the form of KE 7

8 Potential Energy Stored in a Spring Spring at equilibrium Spring is stretched F applied = k x F s = - kx Potential Energy stored in Springs Example 4 (5.7) A 0.50 kg block rests on a horizontal, frictionless surface. The block is presses against a light spring of spring constant k = 80.0 N/m. The spring is compressed a distance of.0 cm to point A and released. Consider the system to be spring, block and Earth. Find a) the speed of the block when it is at the bottom of the incline (position B) b) The maximum distance d the block travels up the incline (θ = 5 0 ) 8

9 Potential Energy stored in Springs Example 4 (5.7) 0 Conservation of mechanical energy (A to B) KE final + PE final =KE initial + PE initial KE B + (PE( gb + PE sb ) =KE A + (PE( ga + PE PE gb PE sb PE ga PE sa ) ½ m v B + (0 +0) = 0 + (0 + ½ kx ) v B = ( k/m ) x v B = 0.5 m/s Elastic PE stored in spring Potential Energy stored in Springs Example 4 (5.7) 0 Conservation of mechanical energy (B to C) KE final + PE final =KE initial + PE initial KE C + PE gc =KE B + PE gb ½ m v B + 0 = 0 + mgh h = dsinθ d = (v B )/(gsinθ) = 7.5x10-3 m 9

10 Potential Energy stored in Springs Example 5 (5.8) A spring (k = 1.00x10 3 N/m) is placed on a table in a vertical position. A block of mass m =1.60 kg is held 1.00 m above the free end of the spring. The block is dropped from rest so that it falls vertically onto the spring. By what distance does the spring compress? Potential Energy stored in Springs Example 5 (5.8) 0 Conservation of mechanical energy (A to B) (system = spring, block, Earth) KE final + PE final =KE initial + PE initial KE B + (PE( gb + PE sb ) =KE A + (PE( ga + PE sa ) PE gb PE sb PE ga PE sa 0 + (0 + ½ kd ) = 0 + (mg(h+d) +0) solve for d (quadratic equation) d = 0.19 m 10

11 Problem 41: Conservation of Energy with Non- Conservative Forces Example 6 A.1x10 3 kg car starts from rest at the top of a 5.0 m long driveway that is sloped at 0 0 with the horizontal. If an average friction force of 4.0x10 3 N impedes the motion, find the speed of the car at the bottom of the driveway. f k N s sin0 0 = h/s 0 h W θ = 0 0 h = s sin0 0 Conservation of Energy with Non- Conservative Forces Example 6 The work done by all non-conservative forces acting within a system equals the change of the total mechanical energy Wnc = (KE( f + PE f ) (KE i + PE i ) (f k cos ) s =(½ mv f + 0) (0 + mg h) (f k cos ) s =(½ mv f + 0) (0 + mg (s sin0 0 ) -f k s + mg s sin0 0 = ½ mv f v f = (gs sin0 0 f k s/m) v f = 3.8 m/s 11

12 Power Ratio of energy transfer to the time interval during which the transfer t occurs P = F v P = W t (average power) F acts on an object moving with an average speed v Units: J/s = Watts (W) = 1 kgm /s 3 1 hp = 550 ft lb/s = 746 W 1 kwh = 3.60x10 6 J kwh is a unit of energy not Power! (Electric bill) Problem 48: Power Example 7 A skier of mass 70 kg is pulled up a slope by a motor-driven cable. a) How much work is required to pull him 60 m up a 30 0 slope (assume frictionless) at a constant speed of.0 m/s? b) What power must a motor have to perform this task? 1

13 a) W nc = KE + PE b) Power = W / t Power Example 7 KE = 0 since the speed is constant, then W nc = mg( y) y = (60 m)sin30 0 = 30 m W nc = mg( y) = (70 kg)(9.80 m/s )(30 m) =.06x10 4 J t = time to travel x = 60 m at v =.0 m/s x = v t t = x / v = (60 m)/(.0 m/s) = 30 s P = (.06x10 4 J)/ (30 s) = 687 J/s = 687 W 1 hp = 746 W 687 W = (687W) (1 hp /746 W) = 0.9 hp 13