<p>1 Work and Energy</p><p>► DEF: Work (W) done by a constant force F exerted on an object through distance d is</p><p> scalar 1J (Joule) = 1Nm</p><p>• • </p><p> Work done by force F is zero if positive negative • the force is exerted but when the force and direction of when the force and direction of no motion is involved: motion are generally in the motion are generally in the no distance moved, no work same directions opposite directions • the force is perpendicular to the force helps the motion force opposes the motion the direction of motion (cos 900 = 0) (for example when normal force is maximum work minimum work perpendicular to the direction of motion, or when force is forcing an object to move in circles) W = Fd W = – Fd</p><p>(the work done by friction force is always negative) </p><p>● Example: Forward force is 200 N. Friction force is 200 N. The distance moved is 200 km. Find the work done by a. forward force F on the car. b. friction force F on the car. c. the work done on the car by the net force. fr 5 F = 200 N; F = 200 N; d = 2 x 10 m fr 0 7 0 7 a. W = Fd cos 0 = 4x10 J b. W = F d cos 180 = – 4 x 10 J F fr fr c. the work done on the car by the net force can be calculated in two ways: 1. Find the work done by each force and add them: W = W + W = 0 F fr 2. Find the net force first and then the work done by net force: W = F d cos θ = 0 (F = 0) net net</p><p>● Example: Mike is cutting the grass using a human-powered lawn mower. He pushes the mower with a force of 45 N directed at an angle of 41° below the horizontal direction. Calculate the work that Mike does on the mower in pushing it 9.1 m across the yard. 0 F = 45 N d = 9.1 m θ = 41 W = Fd cos θ = 310 J</p><p>► Work done by a varying force F - graphically ▪ W = Fd cos θ applies only when the force is constant, but force can vary in magnitude or direction during its action. ▪ Examples: varying force of the golf club on a golf ball 2 ● Example: The lady from the first slide is pulling the car for 2 m with force of 160 N at the angle of 60o , then she gets tired and lowers her arms behind her at an angle of 45o pulling it now with 170 N for next 2 m. Finally seeing the end of the journey she pulls it horizontally with the force of 40 N for 1 m. </p><p> Work done by her on the car is: W = (160 N)(cos 60o)(2m) +(170 N)(cos 45o)(2m) + (40 N)(cos 0o)(1m) W = 80x2 + 120x2 + 40x1 = pink area + green area + blue area = 440 J</p><p>In general: The area under a Force - distance graph equals the work done by that force In physics energy and work are very closely linked. If an object has energy it can do a work. On the other hand the work done on an object is converted into energy. </p><p>● Gravitational potential energy, PE = mg h Minimum force needed to lift up a mass m a vertical distance ∆h at constant speed is F = mg. </p><p>Work done by applied force against gravitational force (when the net force is zero, so there is no acceleration) W = F h cos 00 = mg h</p><p> is stored as gravitational potential energy of the object. PE depends on the reference frame in which it is measured, so we keep in mind some reference level, like desk, lower reservoir of water, ground,…. and potential energy can be calculated relative to different initial positions. It depends only on VERTICAL distance from the initial position, not on the path taken. A ramp can reduce the force – can make life easier The same work is done if the object is pulled along a ramp or vertically upward. ▪ minimum force needed to pull an object along a ramp is F = mg sinθ. 2 0 the work done against gravitational force is: W = F d cos 0 = mg sinθ d = mg h 2 is the same as before and is stored as PE of the object at height h above initial position.</p><p>W = F2 d = F1 h = mgh</p><p>2 ● Kinetic energy, KE = ½ m v</p><p>A moving object possesses the capacity to do work. A moving hammer can be used to do work in driving a nail into a piece of wood. How much work a moving object is capable of doing? The kinetic energy of a body essentially tells us how much work the body is capable of doing. Imagine an object of mass m accelerated from rest by a resultant force, F. After traveling a distance d the mass has velocity v: The work done by net force is:</p><p>W = Fd cos 00 = Fd = mad v2 = u2 + 2ad → ad = v2 /2 so work is W = ½ mv2 </p><p>The work done (by the force F in moving the object a distance d is now expressed in terms of the properties of the body and its motion. The quantity ½ mv2 is called the kinetic energy (KE) of the body and is the energy that a body possesses by virtue of its motion. </p><p>So, ◘ If the net force acting on an body is not zero, then: the change in the kinetic energy of the object is equal to the net work done on the object. </p><p>2 2 W = ∆KE = KEf – KEi = ½ mv – ½ mu ● Work – Kinetic energy relationship ◘ KE depends on the reference frame in which it is measured. When you are sleeping, you have zero KE relative to your bed. ◘ But relative to the sun, you have KE = ½ (60 kg) (30000 m/s)2</p><p>● Units 3</p><p>W = Fd (W) = (1N)(1m) = 1 kg m2//s2 = 1 J PE = mgh (PE) = (1kg)(1m/s2 )(1m) = 1 kg m2//s2 = 1 J 2 2 KE = ½ mv (KE) = (1kg)(1m/s) = 1 kg m2//s2 = 1 J </p><p>● Example: A father pushes his child on a sled on level ice, a distance 5 m from rest, giving a final speed of 2 m/s. If the mass of the child and sled is 30 kg, how much work did he do? What is the average force he exerted on the child? 2 2 W = ∆ KE = ½ m v – 0 = ½ (30 kg)(2) = 60 J</p><p>W = Fd = 60 J, and d = 5 m, so F = 60/5 = 12 N </p><p>● Example: A 1000-kg car going at 45 km/h. The driver is on the cell phone not seeing a tree in front. Upon hitting the tree the car collapses one foot (o.3 m). How much work must this tree/friction force do in order to stop the car? What average force is required to stop a 1000-kg car going at 45 km/h if the car collapses one foot (0.3 m) upon impact? u = 45 km/h = 45 x 1000 m/3600 s = 12.5 m/s 2 2 W = ∆ KE = 0 – ½ m u = – ½ (1000 kg) (12.5 m/s) = – 78125 J </p><p>W = – 78 kJ (the negative sign just means the work leads to a decrease in KE) 3 W = – F d F = – W/d F = 78125 J/0.3 m = 260 x 10 N corresponds to 26 tons hitting you OUCH</p><p>◘ we shall see in nuclear physics that this should actually be the principle of mass-energy conservation ◘ To demonstrate the so-called principle of energy conservation we will solve a dynamics problem in two different ways, one using Newton’s laws and the kinematics equations and the other using the principle of energy conservation.</p><p>● Example: An object of mass 4.0 kg slides from rest without friction down an inclined plane. The plane makes an angle of 30° with the horizontal and the object starts from a vertical height of 0.5 m. Determine the speed of the object when it reaches the bottom of the plane.</p><p>At point A: h = 0.5 m and KE = 0 (u = 0) At point B: h = 0 and the object has gained kinetic energy.</p><p>Method 1: Newton’s law Method 2: Conservation of energy The force down the plane As the object slides down the plane its potential energy becomes transformed into kinetic energy. If we assume that F = mg sinθ no energy is ‘lost ’ (converted into heat), we can write </p><p>PEA = KEB </p><p> v = 3.2 m/s mgh = ½ mv2 v = 3.2 m/s only vertical distance </p><p>● Conservation of energy law ◘ The second solution involves making the assumption that potential energy is transformed into kinetic energy only (no energy is converted into heat, sound….).</p><p>◘ This is ‘energy conservation principle’, this means the energy is conserved. Energy cannot be created or destroyed; it may be transformed from one form into another (for example from potential to kinetic energy), but the total amount of energy doesn’t change. 4</p><p>◘ If it is transformed into thermal energy, we say it is dissipated. The object can’t keep that energy. It is shared with environment. </p><p>◘ When using the energy principle we are only concerned with the initial and final conditions and not with what happens in between. </p><p>◘ No time included in energy principle.</p><p>◘ Very powerful tool of enormous importance. ◘ Clearly in this example it is much quicker to use the energy principle. now problems and examples. Some of them would be very hard to solve using kinematic equations and Newton’s laws. </p><p>◘ Work is a way of transferring energy from one form to another, ● Example: but itself is not a form of energy. All types of energies and work have the same units. Everything can be transformed into each other. ◘ When work is done on an object, energy is transferred to that object. </p><p>● Example: ● Example: A car (toy car – no engine) is at the top of a hill on a A simple pendulum consists of a 2.0 kg mass attached to a string. frictionless track. What must the car’s speed be at It is released from rest at A as shown. Its speed at the top of the first hill if it can just make it to the top the lowest point B is: of the second hill PE + KE = PE + KE A A B B PE + KE = PE + KE 1 1 2 2 2 2 mgh + ½ mv = mgh + ½ mv 1 1 2 2 m cancels out ; v = 0 2 2 gh + ½ v = gh 1 1 2</p><p>● Conservation of energy law AGAIN – WITH FRICTION INCLUDED All previous examples were neglecting friction force, air resistance … What happens if the friction force acts? Can we still use conservation of energy principle?</p><p>● Example: An object of mass 4.0 kg slides 1.0 m down an inclined plane starting from rest. Determine the speed of the object when it reaches the bottom of the plane if</p><p> a. friction is neglected b. constant friction force of 16 N acts on the object as it slides down.</p><p> a. all PE was transformed into KE b. Friction converts part of kinetic energy of the object into heat energy. This energy equals to the work done by the friction. 5</p><p>We say that the frictional force has dissipated energy.</p><p> initial energy = final energy initial energy – Ffr d = final energy (Wfr = – Ffr d)</p><p>PEA = KEB PEA – Ffr d = KEB </p><p>2 2 mgh = ½ mv mgh – Ffr d = ½ mv d = 1.0 m 2 20 – 16 = 2.5 v v = 1.4 m/s</p><p>● Power is the work done in unit time or energy converted in unit time measures how fast work is done or how quickly energy is converted. Power is a scalar quantity.</p><p>Units: </p><p>● Example: A 100 W light bulb converts electrical energy to heat and light at the rate of 100 J every second.</p><p>◘ There is another way to calculate power</p><p>Power is equal to force times velocity, providing that both force and velocity are constant and in the same direction. Constant velocity with a force applied ?????? That’s because we are interested in one force only. Net force is obviously zero. Like power of the engine of the car. </p>