Work And Energy

Work and Energy

Work  Power  Kinetic Energy  Potential Energy  Work and Energy Theorem  Law of Conservation of Energy

Mr. Jayson G. Chavez High School Department Malate Catholic School Malate Manila

Work and Energy WORK • The product of the force and the distance moved by the force. POWER • The time rate of work ENERGY • The property of a system that enables it to do work. MECHANICAL ENERGY • Energy due to the position of something or the movement of something.

Work and Energy POTENTIAL ENERGY • The energy that something possesses because of its position. KINETIC ENERGY • Energy of motion WORK-ENERGY THEOREM • The work done on an object equals the change in kinetic energy of the object.

Work and Energy CONSERVATION OF ENERGY • Energy cannot be created or destroyed, it may be transformed from one form to another, but the total amount of energy never changes.

List of Formula Work = Fd = F cos θ d Power= work/ time Kinetic Energy= ½ mv2 Potential Energy= mgh Work- Energy Theorem Work = KEf – KEi Law of Conservation of Energy ½ mvf2 + mghf = ½ mvi2 + mghi

Problem 7 • Find the work done by a 172.23 N force in pulling a luggage carrier at an angle of 37.21O for a distance of 34.23 m.

Answer 7 (Work) Given: Force= 172.23 N displacement= 34.23 m θ= 37.21o Required: Work done Equation: Work= F cos θ (d) = 172.23 N cos 37.21o (34.23 m) = 172.23 N (0.80) (34.23m) = 137.78 N (34. 23 m) = 4716.21 N  m or simple Work= 4716.21 Joules

Problem 8 • A laborer is lifting a box which has a weight of 213.54 N. He raises the box a distance of 1.35 m above the ground and then lowers the box the same distance. The weight is raised and lowered at a constant velocity. Determine the work done on the box by the worker during a) the lifting phase and b) the lowering phase.

Answer 8 (Work) Given: Weight = 21.54 N Displacement= 1.35 m Required: Work done both lifting phase and lowering phase Equation: Work= F cos θ (d) Lifting phase W= 21.54 N cos 0 (1.35 m) = 21.54 N (1) (1.35 m) W= 29.08 N  m or Joules

Lowering phase W= 21.54 cos 180 (1.35 m) = 21.54 (-1) (1.35 m) W = -29.08 N  m or Joules

Problem 9 • A race car of mass m=1.07 x 103 kg is travelling at a speed of 215 kph through a course. No forces act on the probe except that generated by its own engine. The engine exerts a constant force of 3.5 x 103 N, directed parallel to the displacement. The engine fires continually while the car moves in a straight line for a displacement of 2.25 x 102 m. Determine the final speed of the car.

Answer 9 (Work and Energy Theorem) Given: mass= 1.07 x 103 kg vo= 215 km/hr force= 3.5 x 103 Newtons displacement= 2.25 x 102 m Required: vfinal Equation: STEP 1: Convert the initial velocity to m/s; since, it is expressed in km/hr. 215 km/hr x (1000 m/1km) x (1hr/60min) x (1min/60sec) = 59.72 m/s Note: In 1 km = 1000 m; in 1 hr = 60 min and 1 min= 60 sec

STEP 2: Since there is force and displacement involved you can derive the value for the work done in this situation. W= F cos θ d = 3.5 x 103 N cos 0 (2.25 x 102 m) = 3.5 x 103 N (1) (2.25 x 102 m) = 3.5 x 103 N (2.25 x 102 m) = 787500 Joules or simply W= 7.88 x 105 Joules

STEP 3: Now that you were able to derive the magnitude of the work done by the race car, you can calculate for the vfinal using the formula showing the relationship between work and kinetic energy. W= KEf – KEo KEf = W + KEo = W + ½ mvinitial2 = 7.88 x 105 Joules + ½ 1.07 x 103 kg (59.72 m/s)2 = 7.88 x 105 Joules + 1.91 x 106 Joules KEf= 2.70 x 10 6 Joules Therefore the final kinetic energy is 2.70 x 106 Joules

Cont’d STEP 3: Since we were able to calculate for the value of the final Kinetic energy we can use the formula KEf = ½ mvfinal2 . 2.70 x 106 Joules = ½ 1.07 x 103 kg (vfinal)2 2 (2.70 x 106 Joules) / 1.07 x 103 kg = vf2 5.4 x 106 Joules/ 1.07 x 103 kg = √5046.73 m2 /s2 = vf 71.04 m/s = vf

Answer 9 (Work- Energy Theorem) REASONING: Since the force is parallel to the displacement, the force does positive work on the car. This work which can be calculated directly from the equation W= Fd (Work=7.88 x 105 Joules) produces a change in the Kinetic energy of the space craft. So now we can use the work-energy theorem to find the final kinetic energy (KEf = 2.70 x 106 Joules), and with that we can determine the final speed of the car which is 71.04 m/s .

Problem 10 • A motorcyclist is trying to leap across a course shown in the illustration below by driving horizontally off the cliff. When it leaves the cliff, the motorcycle has a speed of 38.0 m/s. Ignoring air resistance, find the speed with which the cycle strikes the ground on the other side.

H = 70 m

H = 35 m

Answer 10 (Law of Conservation of Energy) Given: vinitial = 38.0 m/s hinitial = 70 m hfinal

= 35 m

Required: vfinal Equation: Note: In this word problem we can use the principle of Law of conservation of energy, since it involves both velocity and height (HINTS).

STEP 1: The law of conservation of energy states that: ½ mvf2 + mghf = ½ mvi2 + mghi Since mass is present in all terms, we can cancel this unit algebraically. ½ vf2 + ghf = ½ vi2 + ghi

You were asked to determine the final velocity (vf) therefore you should manipulate the formula using algebraic concepts. vf2 = vi2 + 2g(ho – hf ) The final velocity in the previous formula is “squared” so in order for you to remove the “square” in vf get the square root of vi2 + 2g(ho – hf ). vf = √vi2 + 2g(ho – hf )

STEP 2: Distribute now the values for vf = √vi2 + 2g(ho – hf ) vf = √[vi2 + 2g(ho – hf )] = √ [(38 m/s)2 + 2 (9.8m/s2) (70m- 35m)] = √ [(1444 m2/s2) + 19.6 m/s2 (35 m)] = √ [(1444 m2/s2) + 686 m2/s2] = √ [2130 m2/s2 ] vf = 46.15 m/s

For Further Readings http://www.physicsclassroom.com/Class/energy/energtoc.html