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7.2 Kinetic and potential energy

7.2 Kinetic and potential energy

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7.2 Kinetic and potential energy

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  1. 7.2 Kinetic and potential energy 7.2 Kinetic and potential energy

  2. 1 Kinetic energy (KE) A force F acts on an object of mass m, increasing its velocity from u tov over a distance s… u v F F s Kinetic energy (KE) gained = work done by force 7.2 Kinetic and potential energy

  3. - v u ( ) 2 2  2a 1 - v u ( ) 2 2 m = 2 1 Kinetic energy (KE) u v F F s KE gained = W=Fs  F=ma v2u2 = 2as & KE gained = Fs =ma 7.2 Kinetic and potential energy

  4. mv2 mu2 1 1 1 1 mv2 - = = mu2 2 2 2 2 1 Kinetic energy (KE) u v F F s KE gained = When the object’s velocity isu KE When the object’s velocity is v KE 7.2 Kinetic and potential energy

  5. 1 Kinetic energy (KE) Simulation 7.2 Kinetic and potential energy

  6. 1 2 70 2 = ( )  1000 3.6 Example 3 KEpassengercar vs KEtruck (a) Find the KE of a passenger car of mass 1000 kgtravelling at (i)70 km h–1, (ii)100 km h–1. (i) KEpassenger car at 70 km h–1 = 189 000 J = 0.189 MJ 7.2 Kinetic and potential energy

  7. 1 2 100 2 = ( )  1000 3.6 Example 3 KEpassengercar vs KEtruck (a) Find the KE of a passenger car of mass 1000 kgtravelling at (i)70 km h–1, (ii)100 km h–1. KE of car at 70 km h1= 0.189 MJ (ii) KEpassenger car at 100 km h–1 = 386 000 J = 0.386 MJ 7.2 Kinetic and potential energy

  8. 1 70 2 = ( )  10 000 2 3.6 Example 3 KEpassengercar vs KEtruck (b)Find the KE of a 10-tonnes truck (10 000 kg)travelling at 70 km h–1. KE of car at 70 km h1= 0.189 MJ KEtruck at 70 km h–1 = 1 890 000 J = 1.89 MJ 7.2 Kinetic and potential energy

  9. 1 1 KE = mv2 2 2 =  1500 202 Example 4 Braking distance (energy approach) A car of mass 1500 kg is travelling at 20 m s–1(72 km h –1). (a)What is the kinetic energy of the car? = 300 000 J = 0.3 MJ 7.2 Kinetic and potential energy

  10. Example 4 Braking distance (energy approach) (b)Brakes are applied to stop the car. If the braking force is 7500 N (equalto half the weight of the car), what is the braking distance? Data: KE of car = 0.3 MJ 7.2 Kinetic and potential energy

  11. 1 1 2 2 mu2 Fs = mv2 Example 4 Braking distance (energy approach) KE of car = 0.3 MJ braking force = 7500 N To stop the car, work done by braking force= KEof the car F = 7500 N force is ve because forward is taken as +ve 7500s = 0  300 000 s = 40 m 7.2 Kinetic and potential energy

  12. 2 Gravitational potential energy (PE) PE is the energy an object possesses due to its position above the ground. mg h A box of mass m is lifted a height h above the ground. Force needed to lift box = weight = mg mg Work done=forcedistance lifted =mgh(PE gained by box) (unit: J) 7.2 Kinetic and potential energy

  13. N.B: • 1. As the height is measured from the ground up, we choose the ground to be the reference point and P.E of the load at the ground level is taken to be zero. • 2. However we could choose some other reference points. Then the potential energy of the load have a different value. But the change in potential energy remains the same with reference to the same reference point. • 3. For the displacement of the load to the level below the reference point, the P.E. of the load is negative. 7.2 Kinetic and potential energy

  14. 2 Gravitational potential energy (PE) Points to note in calculating PE: Only the height lifted against gravity matters, NOT the actual distance moved. less force, but longer distance g massm massm h both have the same PE at the top!! PE = mgh 7.2 Kinetic and potential energy

  15. 2 Gravitational potential energy (PE) Simulation 7.2 Kinetic and potential energy

  16. 1 m 1 m Example 5 Potential energy gain of the box A 2-kg box is lifted 1 m from the floor toa table top and then to a shelf 1 m abovethe table. What is the gravitationalpotential energy the box gains (a) at the table top, (b) on the shelf? 7.2 Kinetic and potential energy

  17. 1 m 1 m Example 5 Potential energy gain of the box (a)PEgain of box at the table top = mgh = 2101= 20 J (b)PEgain of box on the shelf = mgh = 2102= 40 J 7.2 Kinetic and potential energy

  18.  ________ = ____ J  ________ = ____ J mv22 mv12 1 1 1 1 = = = = 2 2 2 2 Q1 A 0.2-kg stone is... A 0.2-kg stone is thrown with 5 m s1.Find thegain in KEof it. 5 m s1 Initial KE (KE1) 0.2 52 2.5 Final KE (KE2) 0.2 302 90 30 m s1 87.5 KE gain = KE2 KE1 = ____ J 7.2 Kinetic and potential energy

  19. Q2 Pauline of mass 50 kg... Pauline of mass 50 kg goes hiking. After walking for5 km, she is 20 m vertically above her starting point. (a)What is her potential energy gained? Potential energy gained = mgh = ____10_____ = _________ J 50 20 10 000 7.2 Kinetic and potential energy

  20. Q2 Pauline of mass 50 kg... (b)If she takes a steeper route and keeps the startand end points the same, will her PEgain be different? Why? Her PE gain will be _______________. It isbecause both her _________________ and______________ displacement are the same. the same weight vertical 7.2 Kinetic and potential energy

  21. The End 7.2 Kinetic and potential energy

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