The roller coaster The Ninja has a height of 122 ft and speed of 52 mph. Its potential energy due to height changes into kinetic energy of motion. Work is done by a force when that force causes an object to move in the direction of the force. The kinetic energy of an object is 1/2mv^2. The potential energy due to gravity is mgh. The work-energy theorem states that the work done on an object is equal to its change in kinetic energy.

1. Work, energy and power

2. The Ninja, a roller coaster at Six Flags over Georgia, has a height of 122 ft and a speed of 52 Mph. The potential energy due to its height changes into kinetic energy of motion.

3. WORK Work is done by force when there is a force applied on the body and the body must move with a displacement in line with the force applied. 𝐹 𝐹 𝐹 𝜃 𝜃 𝜃 𝐹 ∆𝑠 𝜃= angle bet. 𝐹and ∆𝑠 𝐹|| = component of 𝐹 parallel with ∆𝑠 𝜃 ∆𝑠 𝐹|| 𝑊=𝐹||∆𝑠=𝐹∆𝑠cos𝜃 Work done by constant force

4. 𝑾 can be + 𝑖𝑓𝑓 𝐹∥∆𝑠 (0𝑜≤𝜃<90𝑜)− 𝑖𝑓𝑓 𝐹∥∆𝑠 (90𝑜<𝜃≤180𝑜)𝟎 𝑖𝑓𝑓 𝐹 ⊥∆𝑠 (𝜃=90𝑜) Units of work: joule, J (1 J = 1 N-m) erg (1 erg = 1 dyne-cm) ft-lb

5. Example 01 Demi horizontally pushes the 200-N crate in a rough horizontal plane with a constant force of 90 N to continuously move it in uniform motion at a distance of 100 m. What is the total work done on the crate?

6. energy Energyis anything that can be converted into work; i.e., anything that can exert a force through a distance. Energy is the capability for doing work. Unit of energy is the same to the unit of work. Other units used: calorie British Thermal Unit (Btu) kilowatt-hour

7. Kinds of Mechanical Energy Kinetic Energy, K – “speed” Potential Energy, U – “position” or “condition” a. Gravitational PE, Ug b. Elastic PE c. Electric PE Transit Energies: KE and Heat ”

8. Work done and Kinetic Energy 𝑣 𝑣𝑜 𝑃 𝑃 ∆𝑠 𝑎=𝑣2−𝑣𝑜22∆𝑠 𝐹||=𝑚𝑣2−𝑣𝑜22∆𝑠 𝐹=𝑚𝑎 𝐹||∆𝑠=12𝑚𝑣2−12𝑚𝑣𝑜2 𝐾=12𝑚𝑣2 Kinetic energy ∴𝑊=𝐾−𝐾𝑜=∆𝐾 Work-Energy Theorem Work done on the body by resultant forces is its change in kinetic energy

9. Work done by gravity (weight) and gravitational potential energy 𝑊𝑤=𝑤∆𝑠cos𝜃 𝑤 𝑊𝑤=𝑚𝑔𝑦−𝑦𝑜cos180 ∆𝑠 𝑦 𝑊𝑤=𝑚𝑔𝑦𝑜−𝑚𝑔𝑦 𝑤 𝑦𝑜 𝑈𝑔=𝑚𝑔𝑦 Gravitational potential energy ∴𝑊=𝑈𝑜−𝑈=−∆𝑈 Work done on the body is its negative change in potential energy

10. Review first: Work 𝑊=𝐹||∆𝑠=𝐹∆𝑠cos𝜃 𝐾=12𝑚𝑣2 Kinetic Energy Gravitational Potential Energy 𝑈𝑔=𝑚𝑔𝑦 Work-Energy Theorem 𝑊=∆𝐾

11. since 𝑊1+𝑊2+…=∆𝐾 𝑊=∆𝐾 𝑊1=−∆𝑈 𝑊2+ …= 𝑊𝑜𝑡h𝑒𝑟=work done by other forces ∴ −∆𝑈+𝑊𝑜𝑡h𝑒𝑟=∆𝐾 𝑈𝑜−𝑈+𝑊𝑜𝑡h𝑒𝑟=𝐾−𝐾𝑜 Law of Conservation of Energy 𝐾𝑜+𝑈𝑜+𝑊𝑜𝑡h𝑒𝑟=𝐾+𝑈 Initial energy = final energy

12. Examples: Use energy methods to solve all problems A bus slams on brakes to avoid an accident. The thread marks of the tires is 25 m long. If 𝜇𝑘=0.70, what was the speed of the bus before applying brakes? A 1.50-kg book is dropped from a height of 15.0 m from the ground. Find its potential and kinetic energy when it is 6.0 m from the ground. A small rock with a mass of 0.20kg is released from rest at point A, which is at the top edge of a large hemispherical bowl with radius R = 0.80m. Assume that the size of the rock is small in comparison to the radius of the bowl, so the rock can be treated as particle, the work done by the friction when it moves from point A to point B at the bottom of the bowl is -0.22J. What is the speed of the rock when it reaches point B?

13. Power Power is defined as the rate at which work is done. 𝑷=∆𝑾∆𝒕 Power Units of Power: watt, W 1 W=1 J/s erg/s foot=pound per second (ft-lb/s) horsepower 1 hp=746 W

14. Power and velocity Recall average speed or constant velocity: 𝑣=𝑑𝑡 So that 𝑑=𝑣𝑡 Since 𝑊=𝐹𝑑 and 𝑃=𝑊𝑡 𝑃=𝐹𝑑𝑡=𝐹𝑣𝑡𝑡 Power at constant velocity 𝑃=𝐹𝑣

15. Example of Power What power is consumed in lifting a 70.0-kgrobber 1.6m in 0.50 s? 𝑃=𝑊∆𝑡 𝑃=𝐿𝑦𝑡 𝑃=𝑚𝑔𝑦𝑡 𝑃=(70.0 kg)(9.8 ms2)(1.6 m) 0.50 s P= 2200 W=2.2 kW

16. MORE PROBLEMS Use energy methods to solve all problems 1. Tarzan swings on a 30.0-m-long vine initially inclined at an angle of 37.0o with the vertical. What is his speed at the bottom of the swing (a) if he starts from rest? (b) if he pushes off with a speed of 4.00m/s? hint: the work done by tension is zero. 2. A 45.0-kg block of wood initially at rest is pulled by a cord from the bottom of a 27.0o inclined plane. The tension of the cord is 310 N parallel to the plane. After travelling a distance of 2.0 m , the speed of the block is 5.0 m/s. (a) what is the work done by friction? (b) what is the coefficient of friction? 3. A 750-N box is pulled in a rough horizontal plane by a motor driven cable. The coefficient of kinetic friction between the box and the plane is 0.40. (a) How much work is required to pull it 60 m at a constant speed of 2.0 m/s? (b) What power must the motor have to perform this task?