VIEWS: 18 PAGES: 30 POSTED ON: 9/6/2011
Home Work p120: 21 p672: 1 p672: 2 p672: 4 p120: 22 p120: 23 p120: 24 p120: 25 p120: 27 p120: #21) Calculate the work done when a 20 N force pushes a cart 3.5 m? List W = Fd F = 20 N W = 20(3.5) d = 3.5 m W = 70 J W=? p672: #1) Calculate the work done as a 100 N force pushes a crate 5 m along a factory floor? List W = Fd F = 100 N W = 100(5) d=5m W = 500 J W=? p672: #2) Calculate the work done by a person who carries a 50 N box vertically 3 m up a ladder? List W = Fd F = 50 N W = 50(3) d=3m W = 150 J W=? p672: #4) Calculate the power expended by the person who carries the 50 N box vertically 3 m up a ladder in 3 seconds? List P = Fd / t F = 50 N P = 50(3) / 3 d=3m P = 50 W t= 3s P=? p120: #22) Calculate the work done by lifting a 500 N barbell 2.2 m above the floor. What is the potential energy of the barbell when it is lifted to this height? List W = Fd Fg = 500 N W = 500(2.2) W = 1100 J d = 2.2 m W=? PE = mgh = 500(2.2) PE = ? PE = 1100 J p120: #23) Calculate the power expended when the barbell is lifted 2.2 m in 2 seconds. List W = 1100 J P=W/t P = 1100 / 2 t=2s P = 550 W P=? p120: #24)a.) Calculate the work needed to lift a 90 N block of ice a vertical distance of 3m. What PE does it have? List W = Fd Fg = 90 N W = 90(3) = 270 J d=3m PE = mgh = 90(3) W=? PE = 270 J PE = ? p120: #24)b.) When the same block of ice is raised the same vertical distance by pushing it up a 5 m long inclined plane, only 54N of force is needed. Calculate the work needed to push the block up the plane. What PE does it have? List W = Fd F = 54 N W = 54(5) = 270 J d=5m PE = 90(3) = 270 J W=? PE = ? p120: #25) Calculate the change in potential energy of 8 million kg of water dropping 50m over Niagara Falls. List PE = mgh m = 8 000 000 kg PE = 8 000 000(9.8)(50) d = 50 m PE = ? PE = 3 920 000 000 J PE = 3.92 x 109 p120: #26) If 8 million kg of water flows over Niagara Falls each second, what is the power available at the bottom of the falls? List P = W/t m = 8 000 000 kg W = PE d = 50 m P = 3.92 x 109/1 PE = 3.92 x 109 P = 3.92 x 109 W t = 1s P=? p120: #27)a) Calculate the kinetic energy of a 3kg toy cart that moves at 4 m/s. List KE = ½ mv2 m = 3 kg KE = ½ (3)(42) = ½ (3)(16) v = 4 m/s KE = ? KE = 24 J p120: #27)b) Calculate the kinetic energy of the same cart at twice the speed. List KE = ½ mv2 m = 3 kg KE = ½ (3)(82) = ½ (3)(64) v = 8 m/s KE = ? KE = 96 J 12. How much PE does Tim, m = 60 kg, gain when he climbs 3.5 m? List: PE = mgh m = 60 kg PE = 60(9.8)3.5 h = 3.5 m PE = 2058 J PE = ? 13. A 6.4 kg BB is lifted 2.1 m. What is the change in PE? List: PE = mgh m = 6.4 kg PE = 6.4(9.8)2.1 h = 2.1 m PE = 131.7 J PE = ? 14. mary weighs 500 N and walks down 5.5 m. What is the change in PE? List: PE = mgh mg = 500 N PE = 500(-5.5) h = -5.5 m PE = -2750 J PE = ? 15. A WL lifts a 180 kg barbell a height of 1.95 m. What is the change in PE? List: PE = mgh m = 180 kg PE = 180(9.8)1.95 h = 1.95 m PE = 3440 J PE = ? 18. What mass must be lifted a height of 1 m to do 1 J of work? List: PE = mgh m=? 1 = m(9.8)1 h=1m m = 1/9.8 = .102 kg PE = 1 J W=1J Kinetic Energy Kinetic Energy (KE) is the energy of motion. KE = ½ mv2 Example: What is the KE of a 3 kg toy truck moving at 2 m/s? List m = 3 kg KE = ½ mv2 = ½(3)(22) = 6 J v = 2 m/s KE = ? What is the KE of a 3 kg toy truck moving twice that fast, or 4 m/s? List m = 3 kg KE = ½ mv2 = ½(3)(42) = 24 J v = 4 m/s KE = ? In a frictionless system, the kinetic energy of a moving object is equal to the work needed to bring it to that speed. W = KE Fd = ½ mv2 Example: How much work is needed to increase the speed of a 85 kg bicycle from 22 m/s to 46 m/s? List W = ΔKE m = 85 kg W = ½ m(vf2 – vi2) vi = 22 m/s W = ½ (85)(462 – 222) v2 = 46 m/s W = ½ (85)(2116 – 484) W=? W = ½ (85)(1632) = 69360J Class work: p. 672, 673: 5, 6, 7, 8 Show all work 5. Calculate the KE of a 10 kg cart moving at 4 m/s. List: KE = ½ mv2 KE = ½ 10 (42) m = 10 kg KE = 80 J v = 4 m/s KE = ? 6. Calculate the change in KE of a 10 kg cart speeding up from 4 m/s to 10 m/s. List: KE = ½ mv2 KE = ½ 10 (102 - 42) m = 10 kg KE = 5(100 – 16) vi = 4 m/s KE = 5(84) = 420 J vf = 10 m/s KE = ? 7. So it takes 420 J of work to change the KE by 420 J! 8. How much work is done to lift a 3 kg crate up 2 m and throw it at 4 m/s? List: PE = mgh KE = ½ mv2 m = 3 kg PE = 3(9.8)2 KE = ½ 3 (42) v = 4 m/s KE = 58.8 J KE = 24 J h=2m PE = ? total energy = 82.8 J KE = ?