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5. Mechanical Energy Kinetic Energy: the energy of an object in motion depends on mass and velocity E k = 1 2 mv 2 http://www.youtube.com/watch?v=aCdHEWWpWj8&feature=related What is the kinetic energy of a 1000 kg car traveling at 9m/s? E k = 1/2 (1000kg)(9m/s) 2 = 40500J

6. Work Energy theorem Work done = Kinetic Energy (end) -Kinetic Energy (start) W tot = ΔE k = E kf - E ki W tot = F R Δd cos θ Ek = 1 mv 2 2 1/2 mv i 2 1/2 mv f 2 - Wtot = or

7. A car is travelling at 15 m/s, it weighs 600kg What is the kinetic energy? Ek = 1 mv 2 2 Ek = 0.5(600kg)(15m/s) 2 = 67500J

8. p. 344 Example B A dog is pulling a 25.0 kg sled over a horizontal surface of 4.00 m with a force of 50.0N using a horizontal rope. The sled was initially at rest and we assume that friction is negligible. a) What is the total work done? b) What is the sled's final velocity? Data: m= 25.0 kg F = 50.0N θ = 0 0 V i = 0 Δd = 4.0 m Wtot = ? Vf = ? W tot = W FT + W Fg + W FN Wtot = 50.0N 4.0m = 200J Eki = 0 (vi = 0) W tot = Ekf 200J = 1/2m(v f) 2 4 00J/25.0kg= v f 2 v f = 4.00m/s a) b)

9. How much work is needed to accelerate a 1000kg car from 20 to 30 m/s? M = 1000kg vi = 20m/s vf = 30 m/s W=? W = 1/2mv f 2 - 1/2mv i 2 W= 1/2 (1000kg)(30m/s) 2 - 1/2(1000kg)(20m/s) 2 W = 2.5 x 10 5 J

10. Gravitational potential energy E pg = mgh E pg - Gravitational potential energy (J) m mass of object g gravitational acceleration (m/s 2 ) h is height above the reference level (usually the ground) expressed in metres (m) h= 2.0m m =.25kg Epg = mgh =.25kg(9.8m/s 2 )(2.0m)= 4.9J

11. Conservation of mechanical energy Mechanical Energy = Kinetic Energy + Potential Energy Em = Ek + Ep Eki + Epi = Ekf + Epf

12. A 1000kg roller coaster car moves from point A to point B and then to point C What is the gravitational PE at B and C relative to point A? rollercoaster 0m 40m 30m 10m A B C D

13. The law of conservation of total energy states that the total quantity fo energy is constant in an isolated system. If there is friction part of the mechanical energy is transformed into thermal energy. ∴ Eki + Epi = Ekf + Epf + ∆Eth ∆Eth = the final thermal energy or heat

14. Example Estimate the kinetic energy and the velocity required for a 70 kg pole vaulter to pass over a bar 5.0m high. assume the vaulter's height at the level of the bar itself. Δd = 5.0 m - 0.9m = 4.1m 1/2 mv i 2 + 0 = 0 + mgy 2 = (70kg)(9.8m/s 2 )(4.1) = 2.8 x 10 3 J The velocity is 1/2mvi 2 = 2.8 x 10 3 J vi 2 =(2/70kg)2.8 x 10 3 J vi = 8.9 m/s