Work and Energy Notes In this unit our study of motion is going to be approached from the perspective of work and energy.

Work In physics, work is defined as a force acting upon an object to cause a displacement. There are three key words in this definition force, displacement, and cause. In order for a force to qualify as having done work on an object, there must be a displacement and the force must cause the displacement. The formula for work is: W = FdcosƟ where W = work (J, Joules) F = force (N) d = displacement (m) Ɵ = the angle between the force and the displacement vector The cosine theta in the work equation relates to the cause factor it selects the portion of the force which actually causes a displacement.

Example A 2.0 x 102 N force is applied at an angle of 30.0o to the horizontal to move a 10.0 kg object at a constant speed for a distance of 15 m. How much work is done by this applied force? Solution W = FdcosƟ W = (2.0 x 102 N)(15 m)(cos30.0o) W = 2598 J or 2.6 x 103 J Example A rope is used to pull a cart up a ramp as per the diagram below. A force of 245 N is used to pull the cart a distance of 2.5 m. How much work is done?

Solution W = FdcosƟ W = (245 N)(2.5 m)(cos 0o) W = (245 N)(2.5 m)(1) W = 612.5 J or 6.1 x 102 J *we use 0o because F and d in the above scenario are both aligned at the same 28.0o, making the difference between them, Ɵ, equal to zero.

Example A 1.00 x 102 kg crate is being pulled across a horizontal floor by a force F that makes an angle of 30.0o above the horizontal. The coefficient of kinetic friction is 0.200. What should be the magnitude of F so that the net work done by it and the kinetic frictional force is zero? Solution This question combines dynamics with work and energy concepts. Draw a diagram and write a master equation that relates to this scenario.

WF + Wf = 0 Now figure out WF and Wf separately and put back into the above master equation later. WF = FdcosƟ WF = Fdcos30.0o WF = Fd(0.866) WF = 0.866Fd *this will be for master equation To find the work due to friction, Wf, we'll need to know the force due to friction, Ff. But to find Ff, we'll need to know the normal force,

FN. Since the crate is not being lifted off the ground, all the vertical forces must add up to zero: FN FN FN FN FN

+ FsinƟ - Fg = 0 = -FsinƟ + Fg = -FsinƟ + mg = -Fsin30.0o + (1.00 x 102)(10) = -0.500F + 1000

Ff = uFN Ff = (0.200)(-0.500F + 1000) Ff = -0.100F + 200 Wf = -Ffd Wf = -(-0.100F + 200)d Wf = (+0.100F -200)d Wf = 0.100Fd - 200d

*negative; friction is opposite direction *this will be for master equation

WF + Wf = 0 0.866Fd + 0.100Fd - 200d = 0 0.866F + 0.100F - 200 = 0 0.966F -200 = 0 0.966F = 200 0.966F/0.966 = 200/0.966 F = 207 N

*get rid of the d's

Kinetic Energy Consider the case of a net external force F acting on a mass m. This net force is the vector sum of all the external forces acting on the object, and for simplicity, its direction is assumed to be in the same direction as the displacement d. According to Newton's second

law, the net force produces an acceleration a given by a = F/m. Because there is an acceleration, the velocity of the mass increases from vi to vf.

Taking F = ma, multiply both sides by d to obtain: Fd = mad From our study of kinematics we have: vf2 = vi2 + 2ad Solving this equation for ad, we obtain: 2ad = vf2 - vi2 ad = (vf2 - vi2)/2 or: ad = ½vf2 - ½vi2 Substituting this into Fd = mad gives: Fd = mad Fd = m(½vf2 - ½vi2) Fd = ½mvf2 - ½mvi2

In this equation, Fd represents the work done by a net external force. So the formula can become: W = ½mvf2 - ½mvi2 The quantity ½mv2 is called the kinetic energy and the above formula leads to the connection between work and kinetic energy. The work-energy theorem states that when a net external force does work W on an object, the kinetic energy of the object changes from its initial value of KEi to a final value of KEf, the difference between the two values being equal to the work: W = KEf - KEi = ½mvf2 - ½mvi2 Because of the above relationship, energy is also given the unit of Joules (J). In general, the kinetic energy (KE) of an object with mass m and velocity v is given by: KE = ½mv2 Kinetic energy is the energy of motion. Example What is the kinetic energy of a 15.00 kg mass moving with a velocity of 3.50 m/s? Solution KE = ½mv2 KE = 0.5(15.00 kg)(3.50 m/s)2 KE = 91.87 J or 9.19 x 101 J

Example A space probe of mass m = 5.00 x 104 kg is traveling at a velocity of vi = 1.10 x 104 m/s through deep space. No forces act on the probe except that generated by its own engine. The engine exerts a constant external force F of 4.00 x 105 N directed parallel to the displacement. The engine fires continually while the probe makes a straight line for a displacement d of 2.50 x 106 m. Determine the final velocity of the probe. (assume the system is isolated and ignore any gravity) Solution Since the force F is the only force acting on the probe, it is the net external force, and the work it does causes the kinetic energy of the spacecraft to change. The work is positive because F points in the same direction as the displacement d. According to the work-energy theorem, a positive value for W means that the kinetic energy increases. Since the KE increases, we expect the final velocity to be greater than that of the initial velocity. Our master equation becomes: W = KEf - KEi KEf = W + KEi ½mvf2 = Fd + ½mvi2 vf = √[(Fd + ½mvi2)/½m] vf = √[((4.00 x 105)(2.50 x 106) + ½(5.00 x 104)(1.10 x 104)2)/½(5.00 x 104)] vf = 1.27 x 104 m/s Example *This example combines dynamics with work and energy concepts, ie. is challenging! The velocity of a curling rock decreases from 1.35 m/s to 0.98 m/s in coasting 9.5 m across the surface of the ice. Find the coefficient of kinetic friction between the rock and ice.

Solution The work done in slowing down the rock is: W = ½mvf2 - ½mvi2 But because we're being asked for the coefficient of friction, you know we'll have to somehow include : Ff = uFN We don't know the normal force, but we do at least know what gravity is. Therefore, change this to: Ff = umg It is the friction that does the work. Also the force of friction is opposite to the direction of motion and therefore needs to be negative: W = -Ffd W = -umgd Lastly, since W = ½mvf2 - ½mvi2 and write: ½mvf2 - ½mvi2 = -umgd ½vf2 - ½vi2 = -ugd Now solve for u: 2 ½vf - ½vi2 = -ugd -ugd = ½vf2 - ½vi2 u = [½vf2 – ½vi2]/-gd u = [½(0.98)2 – ½(1.35)2]/-(10)(9.5) u = 0.00453 or 4.5 x 10-3

W = -umgd, we can

***In-Class Question 10

Gravitational Potential Energy Kinetic energy, energy of motion, is not the only type of energy. Potential energy is the stored energy of position. An example of such energy can be found by an object having a position above the surface of the Earth. Gravitational potential energy, PE, is the energy that an object of mass m has by virtue of its position relative to the surface of the Earth (or some arbitrary zero position) measured by the height h of the object. As a formula: PE = mgh Example The CN tower in Toronto is advertised as being the world's tallest free standing structure. At 181 stories, it has a height of 553 m. What is the gravitational potential energy of a 55.0 kg person who is at the top of the tower? Solution PE = mgh PE = (55.0)(10)(553) PE = 3.04 x 105 J When an object is lifted up work is done when the object is moved from one position to a higher one. Our work formula can become: W = mghf - mghi Example How much work is done when a 5.0 kg object is moved from a height of 0.54 m to one of 0.97 m?

Solution W = mghf - mghi W = (5.0)(10)(0.97) - (5.0)(10)(0.54) W = 22 J Another type of potential energy is elastic potential energy but is not covered in this course.

Conservative and Non-Conservative Forces Earlier we spoke of the work-energy theorem and it stated that when a net external force does work on an object, the energy of the object changes. In order to better understand this idea, it is important to understand conservative and non-conservative forces. Gravity is an example of a conservative force. The work done by gravity does not depend on the choice of path. For instance, when an object is moved from an initial height hi to a final height hf, the object could be raised straight up or it could be moved along any curved path from the initial height to the final height. Any side to side motion does not affect the final work done by gravity. It is only the difference in height that is important. This example leads to one possible version of the definition of a conservative force. Definition 1: A force is conservative when the work it does on a moving object is independent of the path of the motion between the object's initial and final positions. Other examples of conservative forces are the elastic force of a spring and the electrical force. There is another way to view the idea of a conservative force. Imagine a roller coaster starting at the top of its motion. The roller coaster car races through dips and double dips, and ultimately returns

to its starting point. This kind of path, which begins and ends at the same place, is called a closed path. If we assume that there is no friction or air resistance, then gravity is the only force that does work on the car. The track does exert a normal force, but this force is always directed perpendicular to the motion, and hence does no work. On the downward parts of the trip, the gravitational force does positive work, increasing the car`s kinetic energy. On the upward parts of the motion, gravitational force does negative work, decreasing the car`s kinetic energy. Over the entire trip, the gravitational force does as much positive work as negative work, so the net work is zero. The car returns to its starting point with the same kinetic energy it had at the start. This example leads to the other possible version of the definition of a conservative force. Definition 2: A force is conservative when it does no work on an object moving around a closed path, starting and finishing at the same point. We must keep in mind this second version of the definition of conservative forces when working with the upcoming formula.

Conservation of Mechanical Energy The work-energy theorem has led us to consider two types of energy: kinetic energy and potential energy. An object at any given time can have one or even both of these energy types. We can thus speak of an object`s total mechanical energy, E, and this is simply the sum of these two energy types: E = KE + PE Recall that when either of these two energy types change, work must have been done. In terms of the work done, our formula now becomes:

W = (KEf - KEi) + (PEf - PEi) or W = (KEf + PEf) - (KEi + PEi)

The expression (KEf + PEf) is the final mechanical energy, Ef The expression (KEi + PEi) is the initial mechanical energy, Ei. Take W = (KEf + PEf) - (KEi + PEi) and apply our second definition of conservative forces. The definition said a force is conservative when it does no work on an object...and since we will be only considering closed system examples (no friction or air resistance), no work will be done by any outside forces. Suddenly W = (KEf + PEf) - (KEi + PEi) becomes: 0 = (KEf + PEf) - (KEi + PEi) or (KEi + PEi) = (KEf + PEf) Example One of the fastest roller coasters in the world is the Magnum XL-200 at Cedar Point Park in Sandusky, Ohio. The ride includes a vertical drop of 59.4 m. Assume that the coaster has a speed of nearly zero as it crests the top of the hill. Neglect friction and air resistance and find the speed of the riders at the bottom of the hill. Solution (KEi + PEi) = (KEf + PEf) ½mvi2 + mghi = ½mvf2 + mghf ½mvf2 + mghf = ½mvi2 + mghi ½vf2 + ghf = ½vi2 + ghi vf = √[(½vi2 + ghi – ghf)/½] vf = √[(½(0.00)2 + (10)(59.4) – (10)(0.00)/½]

vf = 3.45 x 101 m/s

Power Work is performed when a force causes a displacement. But there is no consideration of time in this simple definition. The work could have been done quickly or slowly. Power is the rate at which work is done. As an equation: P = W/t where P = power (W, Watts) W = work (J) t = time (s) For the same amount of work, power and time are inversely proportional. For example, a more powerful engine can do the same amount of work of a less powerful engine, but in less time. Another formula can be derived from Fd, we can have: P = W/t P = Fd/t Lastly, since velocity is d/t, we can have: P = Fd/t P = Fv

P = W/t. Since W =

This last equation for power tells us that a powerful machine can be strong (big force) or fast (big velocity), or maybe even both at once. ***In-Class Questions 11, 12 Interesting fact: Émilie du Châtelet (1706 - 1749) was a French mathematician, physicist, and writer during the years of history called the Age of Enlightenment. While her most famous achievement is considered to be her translation of Isaac Newton's Principia Mathematica, she gained some notoriety by daring to question some of Newton's ideas on kinetic energy. In her book called Institutions de Physique she combined the theories of Gottfried Leibniz and the observations of Willem 's Gravesande to show that the energy of a moving object is proportional not to its velocity, as had previously been believed by Newton, Voltaire and others, but to the square of its velocity (KE = ½mv2). She was correct, but it took nearly 100 years after her death for this idea to be accepted, no doubt due to the sexist attitudes of the day. Voltaire wrote, considered a compliment at the time, that "She was a great man, whose only fault was being a woman." ***Unit Questions