1. Work, Energy, and Energy Conservation Chapter 5, Sections 1 - 3 Pg. 168-186

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

3. 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)

4. 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

5. 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.

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

7. 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

8. 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

9. 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

10. 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

11. 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

12. 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

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

14. 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

15. 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

16. 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.