Chapter 6 Work & Energy

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Kinetic Energy, E k = 1/2m v 2 (Joule, J) 4. otential Energy, E = m g h (Joule, J) 5.

1 Chapter 6 Work & Energy In physics, work is done by forces on an object over a distance. 1a. W = F d 1b. W = F d Cosθ (Joule, J) 2. W net = E 5. ower, = W t (Watts, W) ext 3. Kinetic Energy, E k = 1/2m v 2 (Joule, J) 4. otential Energy, E = m g h (Joule, J) 5. pring, E = ½kx 2 (Joule, J) Force is at an angle to distance Force is parallel to distance

2 Warm Up (11/20/15) ooke s Law Draw Deformation the diagram below. of a pring It illustrates ooke s Law. Natural length Compression Extension x = 0 -x F sp +x F sp The deformation (x) of a spring is measured relative to its unstretched length. Force (N) F sp = k x pring Constant (N/m) Deformation (m)

3 Lab 10 Determining the pring Constant, k F sp = k x 1 Weight = m g = F w x2 x1

4 Chapter 6 Work & Energy Guiding Questions ow is the energy of a system defined? ow is work represented graphically? What is mechanical energy and what factors affect its conservation?

5 Warm Up (12/02/15) Types of Energy Describe the types of energy present in each photo burning t chemical oxygen in (light ene

6 Types of Energy Energy can be either kinetic or potential. x2 x1

7 Mechanical Energy Mechanical energy is not always conserved, but total energy is. The work done by the force of friction as the piston expands and compresses inside the cylinder generates heat.

8 Conservation of Energy

9 Work Changes Energy 50. kg sled is pulled for 5.2 m with a horizontal force of 100 N, starting from rest. Find the final speed. Ignore friction. W net, ext = E; system: sled, m = 50kg 0 0 because θ = 90 W Fg + W FN + W F = KE; one body system can only have KE 1, θ = FdCosθ = (KE f KE i ) v f = Fd = ½mv f 2 2Fd m = 2(100N)(5.2m) 50 kg v f = 4.6 m/s 0, from rest!

10 Work Changes Energy 50. kg sled is pulled for 5.2 m with a horizontal force of 100 N, starting from rest. Find the final speed. Ignore friction. W net, ext = E; system: sled, m = 50kg 0 0 because θ = 90 W Fg + W FN + W F = KE; one body system can only have KE 1, θ = FdCosθ = (KE f KE i ) v f = Fd = ½mv f 2 2Fd m = 2(100N)(5.2m) 50 kg v f = 4.6 m/s 0, from rest!

11 Daily Quiz #5 Work & Energy 1500 kg car is brought to a stop by a force of 7890 N in a distance of 20 m. ow fast was the car moving?

12 1,200 kg car rolls down a hill with its brakes off and transmission in neutral. t one moment it is moving 5.0 m/s. little later it is moving at 8.2 m/s. ow much did its height change between the two times? ssume friction and air resistance have no effect. Wnet, ext = E; system: car, m = 1200kg 0 because θ = 90 WFg + WFN = KE Δh s 0 + FgdCos60 = (KEf KEi) Fgd( h s) = ½mvf 2 ½mvi 2 h = ½[ (8.2m/s) 2 (5m/s) 2 ] 9.8m/s 2 h = 2.2 m h = ½mvf 2 ½mvi 2 F g h = ½m(vf 2 vi 2 ) mg 60 s

13 Work Energy ractice problem 1,200 kg car rolls down a hill with its brakes off and transmission in neutral. t one moment it is moving 5.0 m/s. little later it is moving at 8.2 m/s. ow much did its height change between the two times? ssume friction and air resistance have no effect. W net, ext = E; system: car, m = 1200kg & Earth 0 because θ = 90 W FN = E 0 = E In the absence of non-conservative forces, mechanical energy is conserved. force is conservative if the work it does on a particle depends only on the initial and final positions of the particle. Conservative forces F g & F sp = -kx Non-conservative force F fr Fg is not an external force

14 Work Energy ractice problem 1,200 kg car rolls down a hill with its brakes off and transmission in neutral. t one moment it is moving 5.0 m/s. little later it is moving at 8.2 m/s. ow much did its height change between the two times? ssume friction and air resistance have no effect. W net, ext = E; system: car, m = 1200kg & Earth 0 because θ = 90 W h = ½[ (8.2m/s) 2 (5m/s) 2 FN = E ] 0 = E In the absence of non-conservative forces, mechanical energy is conserved. 0 = (KE f KE i ) mg h h = ½m(v f 2 v i 2) mg E decreased h = 2.2 m 9.8m/s 2 Fg is not an external force

15 Work Changes Energy uppose the spring and block are oriented vertically, as shown in the diagram. Initially, the spring is compressed 4.60 cm and the block is at rest. When the block is released, it accelerates upward. Find the speed of the block when the spring has returned to its equilibrium position. W net, ext = E; system: block, Earth, pring 0 = E F sp m = 1.70 kg F w 0 = KE + E g + E sp 0, from rest! 0 = (KE f KE i ) + mg h ½kx 2 0 = ½mv f 2 + mg h ½kx 2 k = 955 N/m

Work Changes Energy W net, ext = E; system: block, Earth, pring uppose the spring and block are oriented vertically, 0 = as Eshown in the diagram. Initially, the spring is compressed 4.

16 Work Changes Energy W net, ext = E; system: block, Earth, pring uppose the spring and block are oriented vertically, 0 = as Eshown in the diagram. Initially, the spring is compressed 4.60 cm and the block is at 0 = KE + E g + E sp rest. When the 0, block from rest! is released, it accelerates 0 upward. = (KE f Find KE i ) the + mg h speed of ½kx the 2 block when the spring has returned to its equilibrium position. 0 = ½mv f 2 + mg h ½kx 2 ½mv 2 f = mg h + ½kx 2 v 2 f = mgd + ½kx 2 ½m v f = m/s F sp F w v f = mgd + ½kx 2 v f = ½k k = 1.70kg(9.8m/s2)(0.046m) + ½(955N/m)(0.046m) 2 ½(1.70 kg)

Physics 101 Lecture 7 Kinetic Energy and Work

Physics 101 Lecture 7 Kinetic Energy and Work Phsics 101 Lecture 7 Kinetic Energ and Work Dr. Ali ÖVGÜN EMU Phsics Department www.aovgun.com Wh Energ? q Wh do we need a concept of energ? q The energ approach to describing motion is particularl useful