2Recall:A change in an object’s momentum is related to the amount of time a force acts on an object.Now we will look at the effect of a force acting on an object over a distance.
3WorkThe product of a force exerted on an object and the displacement of the object as a result of the force
4When is work done on an object? 1. a force is applied And when 2. object moves in the direction that the force is applied
5Ex) work is done when an object is lifted against gravity The greater the mass of the object, the more work is done- If an object does not move in the direction of the force, no work is done
6Two categories1. Work done against another force Ex) archer stretches a bowstring (work done against elastic forces in bow) 2. Work done to change the speed of an object Ex) bringing an automobile up to speed on a highway **Simplest case: force is constant and motion takes place in a straight line
7Formula:Work = Force x distanceW = F dUnits: 1 Joule = 1 N x 1 m
8The third floor of a house is 8 m above street level The third floor of a house is 8 m above street level. How much work is needed to move a 150 kg refrigerator to the third floor?
9Determine the work done by a 45 Determine the work done by a 45.0 N force in pulling a suitcase at an angle of 50.0⁰ for a distance of 75.0 m.
10Power Definition: the rate at which work is done Formula: Power = work done / time intervalUnit: Watt = Joule / second1 watt of power is expended when 1 Joule of work is done in one second
11PowerHigh power engines don’t necessarily do more work, they just do the same amount of work in less time.An engine that has 2x the power of a smaller engine will do the same amount of work in half the time.
12A small electric motor is used to lift a 0 A small electric motor is used to lift a 0.50-kilogram mass at constant speed. If the mass is lifted a vertical distance of 1.5 meters in 5.0 seconds, what is the average power developed by the motor?
13Energy – the ability to do work Types of energy: mechanical thermal electric electromagnetic nuclear chemical
14Mechanical Energy Kinetic Energy energy of motion KE = ½ m v2 Potential Energy Stored energy or Energy of position g PE = m g h
15Work –Energy TheoremMoving objects can do work on another object that it comes into contact withA moving object exerts a force on a second object and moves it a distanceTherefore, an object in motion has the ability to do work and can be said to have energy (KE)
16Calculations using the work-energy theorem The work done on an object is equal to its change in KEW = D KEW = KEf – KEiW = ½ m vf2 – ½ m vi2if positive work is done on an object, the KE increases by the work doneif negative work is done, the KE of the object decreases by the work done (force acts in the opposite direction of motion)
17If a 1490 kg car at rest is pushed a distance of 25 m until it reaches a speed of 2.0 m/s. What was the car’s change in kinetic energy?DKE = 2980 JWhat was the work done on the car by the individual pushing the car?W = 2980 JWhat was the average force supplied by the pusher over the 25 m?F = N
18Kingda Ka accelerates riders from rest to 57.2 m/s over 100 m. What is the change in KE of an 80.0 kg student, while the ride accelerates him?How much work does the seat do on the rider?What is the average force of the seat on the rider?
19Types of Potential Energy Gravitational PE2. Elastic PE
20Gravitational Potential Energy (gPE) an object can have PE because of position relative to Earthobject has the ability to do work as a result of it fallingin order to lift a mass, m, vertically a force at least equal to its weight must be exerted on itthe work done lifting the object vertically is equal to the PE of the objectW = DPEW = Force (Weight) x distance (height)W = (m x g) x (h)the gPE is dependent on the vertical height measured from some reference point (floor, sea level)the work an object can do when it falls does not depend on the path taken, but only on the vertical height (free fall; down an incline)
21Elastic Potential (springs, rubber bands, bungee cords) 1. PE associated with elastic substancesEx) when an object is stretched or compressed it has the potential to do work when returning to its original (equilibrium) lengthThe change in length of stretch or compression (x) is proportional to the force exerted (F) to do the stretching or compressingFs = kx k = spring constantk measures the stiffness of the elastic substanceThe spring itself exerts a restoring force in the opposite direction of the displacement.
22Elastic PotentialStretched or compressed elastic objects have potential energy as a result of their ability to do exert a restoring force. The PE stored in the spring can be determined from the following: PEs = ½ k x2
23Energy ConversionsRecall: The work done by all forces acting on an object is equal to the total change in Kinetic and Potential energy on the object W = DKE + DPE
24Energy Transformations: When dealing with mechanical energy we typically deal with Kinetic and PotentialEnergy can be transformed from one form to anotherEx) a falling object decreases its PE, yet its velocity increases constantly, therefore its KE increasesWater falling at Niagara Falls has PE, while it falls the KE is used to turn turbines, which are used to convert mechanical energy to electricalWork is done when energy is transferred from one object to another.
25Conservation of Energy **Whenever energy is transformed, no energy is lost or gained in the process.Consider a falling object, neglecting air resistance, of course:Before being dropped the PE is mgh (with reference to some point)as it falls the height of the object decreases,therefore PE must decreasealso as it falls, the object accelerates to ground,so velocity is increasingtherefore KE must also be increasing as given by ½ mv2
26Conservation of Energy (continued) Just before our object hits the ground, all of the PE is converted to KESo KEbottom = PEtop
27Law of Conservation of Energy The total energy is neither increased nor decreased in any process.Energy can be transformed form one form to another, and transferred from one object to another, but the total energy remains constant.KE + PE = constantORTotal E at pt 1 = Total E at pt 2
28The diagram below shows a 0 The diagram below shows a 0.1-kilogram apple attached to a branch of a tree 2 meters above a spring on the ground below.The apple falls and hits the spring, compressing it 0.1 meter from its rest position. If all of the gravitational potential energy of the apple on the tree is transferred to the spring when it is compressed, what is the spring constant of this spring?Answer : k = N/m
29Mechanical Energy (with friction) In real world problems as potential energy is converted to kinetic energy, and vice versa, friction can do work to impede the energy conversion. Some mechanical energy can be converted to heat energy (Q) during the conversion. When friction is considered we use the following equation: ET = PE + KE + Q
30A roller coaster car has a potential energy of 450,000 J and a kinetic energy of 120,000 J at point A along its track. At the lowest point in the ride the potential energy is zero. And 50,000 J of work have been done by friction after it leaves point A. What is the kinetic energy of the car at its lowest point?ET = PE + KE + Q 450,000 J + 120,000 J = 0 J + KE + 50,000 J 520,000 J = KE
31KEtop + PEtop = KEbottom + PEbottom A 420 N child sits on a swing that hangs 0.40 m above the ground. Her mother pulls the swing back and releases her when the seat is 1.5 m above the ground.How fast is she moving when the swing passes through its lowest point?KEtop + PEtop = KEbottom + PEbottom½ mv2 + mgDh = ½ mv2 + mgh(420 N)( m) = ½ (42.8 kg)(v2)v = 4.6 m/s
32(420 N)(1.5m – 0.4 m) = ½ (42.8 kg)(3.5 m/s)2 + Q b. If she is moving through the lowest point at 3.5 m/s, how much energy was lost to friction?ET = PE + KE + Q(mgDh)top = 0 J + ½ mv2 + Q(420 N)(1.5m – 0.4 m) = ½ (42.8 kg)(3.5 m/s)2 + Q200 J = Q
33mghtop = mghbottom + ½ mv2bottom + Q A roller coaster car has a mass of 290. kilograms. Starting from rest, the car acquires 3.13 × 105 joules of kinetic energy as it descends to the bottom of a 115 m hill in 5.3 seconds.Calculate the energy “lost” do to friction as the roller coaster went down the hill.ET = PE + KE + Qmghtop = mghbottom + ½ mv2bottom + Q(290 kg)(9.81m/s2)(115 m) = 0 J x 105 J + Q14,200 J = Q
34KE = ½ mv2 3.13 x 105 J = ½ (290 kg)(v2) 46.5 m/s = v Calculate the speed of the roller coaster car at the bottom of the hill.KE = ½ mv x 105 J = ½ (290 kg)(v2) 46.5 m/s = v
35a = Dv/t a = (46.5 m/s – 0 m/s) / 5.3 s a = 8.8 m/s Calculate the magnitude of the average acceleration of the roller coaster car as it descendsa = Dv/t a = (46.5 m/s – 0 m/s) / 5.3 s a = 8.8 m/s