2. 1 7.2 Kinetic and potential energy

3. 2 1Kinetic energy (KE) A force F acts on an object of mass m, increasing its velocity from u to v over a distance s… Kinetic energy (KE) gained F s = work done by force F u v

4. 3 7.2 Kinetic and potential energy 1Kinetic energy (KE) F s F u v KE gained =W = FsW = Fs  F = ma &  KE gained = Fs = ma 2a2a uv 22   )( 2 uv 22  )( 1 m = v 2  u 2 = 2as

5. 4 7.2 Kinetic and potential energy 1Kinetic energy (KE) KE gained = mv 2  2 1 mu2mu2 2 1 F s F u v When the object’s velocity is u KE mu2mu2 2 1 = When the object’s velocity is v KE mv2mv2 2 1 =

6. 5 7.2 Kinetic and potential energy 1Kinetic energy (KE) Simulation

7. 6 7.2 Kinetic and potential energy (a)Find the KE of a passenger car of mass 1000 kg travelling at (i) 70 km h –1, (ii) 100 km h –1. (i) KE passenger car at 70 km h –1 = 0.189 MJ = 189 000 J Example 3 KE passenger car vs KE truck  1000  2 1 = ( 3.6 70 2 )

8. 7 7.2 Kinetic and potential energy (a)Find the KE of a passenger car of mass 1000 kg travelling at (i) 70 km h –1, (ii) 100 km h –1. (ii) KE passenger car at 100 km h –1 = 0.386 MJ = 386 000 J Example 3 KE passenger car vs KE truck  1000  2 1 = ( 3.6 100 2 ) KE of car at 70 km h  1 = 0.189 MJ

9. 8 7.2 Kinetic and potential energy (b)Find the KE of a 10-tonnes truck (10 000 kg) travelling at 70 km h –1. Example 3 KE passenger car vs KE truck KE truck at 70 km h –1 = 1.89 MJ = 1 890 000 J  10 000  2 1 = ( 3.6 70 2 ) KE of car at 70 km h  1 = 0.189 MJ

10. 9 7.2 Kinetic and potential energy A car of mass 1500 kg is travelling at 20 m s –1 (72 km h –1 ). = 300 000 J = 0.3 MJ Example 4 Braking distance (energy approach) (a)What is the kinetic energy of the car? KE = mv2mv2 2 1  1500  20 2 2 1 =

11. 10 7.2 Kinetic and potential energy (b)Brakes are applied to stop the car. If the braking force is 7500 N (equal to half the weight of the car), what is the braking distance? Example 4 Braking distance (energy approach) Data: KE of car = 0.3 MJ

12. 11 7.2 Kinetic and potential energy To stop the car,  7500  s = 0  300 000 s = 40 m Example 4 Braking distance (energy approach) KE of car = 0.3 MJ braking force = 7500 N F = 7500 N work done by braking force = KE of the car Fs = mv2 mv2  2 1 2 1 mu2mu2 force is  ve because forward is taken as +ve

13. 12 7.2 Kinetic and potential energy 2Gravitational potential energy (PE) PE is the energy an object possesses due to its position above the ground. A box of mass m is lifted a height h above the ground. Force needed to lift box = weight Work done = force  distance lifted = mgh (PE gained by box) (unit: J) mgmg mgmg h = mg

14. 13 7.2 Kinetic and potential energy 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.

15. 14 7.2 Kinetic and potential energy Points to note in calculating PE: Only the height lifted against gravity matters, NOT the actual distance moved. h PE = mgh both have the same PE at the top!! less force, but longer distance mass m g 2Gravitational potential energy (PE)

16. 15 7.2 Kinetic and potential energy 2Gravitational potential energy (PE) Simulation

17. 16 7.2 Kinetic and potential energy A 2-kg box is lifted 1 m from the floor to a table top and then to a shelf 1 m above the table. What is the gravitational potential energy the box gains Example 5 Potential energy gain of the box (a) at the table top, (b) on the shelf? 1 m

18. 17 7.2 Kinetic and potential energy (a)PE gain of box at the table top (b) PE gain of box on the shelf Example 5 Potential energy gain of the box = 2  10  1 = 20 J = mgh = 2  10  2 = 40 J 1 m

19. 18 7.2 Kinetic and potential energy A 0.2-kg stone is thrown with 5 m s  1. Find the gain in KE of it. Initial KE ( KE 1 ) Q1 A 0.2-kg stone is... 5 m s  1 mv12mv12 2 1 =  ____  ____ = ____ J 2 1 = 0.25252 2.5 Final KE ( KE 2 ) mv22mv22 2 1 =  ____  ____ = ____ J 2 1 = 0.230 2 90 30 m s  1 KE gain = KE 2  KE 1 = ____ J 87.5

20. 19 7.2 Kinetic and potential energy Pauline of mass 50 kg goes hiking. After walking for 5 km, she is 20 m vertically above her starting point. Potential energy gained = mgh = ____  10  _____ = _________ J 502020 10 000 Q2Pauline of mass 50 kg... (a)What is her potential energy gained?

21. 20 7.2 Kinetic and potential energy (b)If she takes a steeper route and keeps the start and end points the same, will her PE gain be different? Why? Her PE gain will be _______________. It is because both her _________________ and ______________ displacement are the same. the same weight vertical Q2Pauline of mass 50 kg...

22. 21 7.2 Kinetic and potential energy The End

23. 22 7.2 Kinetic and potential energy Return