1. Energy and Energy Conservation

2. Energy Two types of Energy: 1. Kinetic Energy (KE) - energy of an object due to its motion 2. Potential Energy (PE) - energy associated with an object due to the position of the object. Energy is the ability to do work.

3. Kinetic Energy Kinetic energy depends on the speed and the mass of the object. KE = ½ mv²

4. Sample Problem What is the kinetic energy of a 0.15 kg baseball moving at a speed of 38.8 m/s? KE = ½ mv² KE = (½)(0.15 kg)(38.8m/s)² KE = 113 J

5. Work-Kinetic Energy Theorem The net work done on an object is equal to the change in kinetic energy of an object. W net = ΔKE W net = ½mv f ² - ½mv i ²

6. Sample Problem A 50 kg sled is being pulled horizontally across an icy surface. After being pulled 15 m starting from rest, it’s speed is 4.0 m/s. What is the net force acting on the sled? W net = ½mv f ² - ½mv i ² V i = 0 m/sV f = 4.0 m/s ∑Fd = ½mv f ² ∑F = (½mv f ²)/d = [(½)(50kg)(4.0m/s)²]/ 15 m ∑F ≈ 27 N

7. Potential Energy Potential energy (PE) is often referred to as stored energy. Gravitational potential energy (PE g ) depends on the height (h) of the object relative to the ground. PE g = mgh

8. Sample Problem What is the gravitational potential energy of a 0.25 kg water balloon at a height of 12.0 m? PE g = mgh PE g = (0.25 kg)(9.81 m/s²)(12.0 m) PE g = 29.4 J

9. Potential Energy Elastic potential energy (PE elastic ) is the potential energy in a stretched or compressed elastic object. PE elastic = ½ kx² “X” is referred to as the distance of the spring compressed or stretched. “K” is the spring constant and is expressed in N/m.

10. Sample Problem Calculate the elastic potential energy of a block spring, with a spring constant of 2.3 N/m, that has a compressed length of 0.15 m and a maximum stretch length of 0.55 m? PE elastic = ½ kx² x = 0.55 m – 0.15 m= 0.40 m = ½ (2.3 N/m)(0.40 m)² PE elastic = 0.18 J

11. Conservation of Mechanical Energy Law of conservation of energy: Energy is neither created or destroyed. It simply changes form. Mechanical energy (ME) is the sum of kinetic and all forms of potential energy. ME = KE +∑PE

12. h 100 % PE 0 % KE 50 % PE 50 % KE 0 % PE 100 % KE Total mechanical energy remains constant in the absence of friction.

13. Sample Problem Starting from rest, a child zooms down a frictionless slide from an initial height of 3.00 m. What is the child’s speed at the bottom of the slide? The child’s mass is 25.0 kg. h i = 3.00 m m = 25.0 kg h f = 0 m v i = 0 m/s v f = ? m/s

14. ½ mv i ² + mgh i = ½ mv f ² +mgh f (25.0 kg) (9.81 m/s²) (3.00 m) = (½)(25.0 kg) (V f )² 736 J / (12.5 kg) = V f ² Vf ² = 58.9 m²/s² Vf = 7.67 m/s h i = 3.00 m m = 25.0 kg h f = 0 m v i = 0 m/s v f = ? m/s

15. Mechanical Energy in the presence of friction In the presence of friction, measured energy values at start and end points will differ. fF app KE Total energy, however, will remain conserved.

16. Work

17. Work W= ∑ Fd(cos θ) Any force that causes a displacement on an object does work (W) on that object. ΣFΣF d

18. Work is done only when components of a force are parallel to a displacement. W= ∑ Fd(cos θ) Work F θ d ΣFΣF Work is expressed in Newton meters (Nm) = Joules (J)

19. Sample Problem How much work is done on a box pulled 3.0 m by a force of 50.0 N at an angle of 30.0° above the horizontal? W= ∑ Fd(cos θ) 50.0 N 30.0° d ΣFΣF = (50.0 N x 3.0 m)(cos 30.0°) W = 130 J

20. Efficiency is a measure of how much of the work put into a machine is changed into useful work by the machine. Efficiency = (W out /W in ) x 100 %Efficiency

21. Sample Problem A man expends 200 J of work to move a box up an inclined plane. The amount of work produced is 40 J. What is the efficiency of the inclined plane? Efficiency = (W out /W in ) x 100 % = (40 J/ 200 J) x 100 = 20 %

22. Momentum and Impulse

23. Momentum and Impulse Momentum and Impulse Momentum is a measure on how difficult it is to stop a moving object. Momentum is a vector quantity. p = mν Measured in kg m/s

24. Objects with a high momentum can have a greater mass, velocity, or both! 1 2 ν 1 = ν 2 m 1 > m 2

25. Falling Object 1Falling Object 2 ν 1 > ν 2 m 1 = m 2

26. A change in momentum takes force and time. This product of force and the time over which it acts on an object is known as an impulse (FΔt). FΔt = mv f – mv i Impulse-Momentum Theorem FΔt = Δp

27. Wall exerts an impulse on the moving ball, thereby causing a change in momentum p1p1 p2p2

28. Sample Problem A 1400 kg car moving westward with a velocity of 15 m/s collides with a pole and is brought to rest in 0.30 s. What is the magnitude of the force exerted on the car during the collision? (Pg. 211)

29. m = 1400 kg ν i = -15 m/s FΔt = mν f – mν i Δt = 0.30 s ν f = 0 m/s F (0.30 s) = (1400 kg)(0 m/s) – (1400 kg)(- 15 m/s) F (0.30 s) = 21,000 kg m/s F = 7.0 x 10 4 N to the East