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Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley PowerPoint ® Lectures for University Physics, Twelfth Edition – Hugh D. Young and Roger A. Freedman Lectures by James Pazun Chapter 6 Work and Kinetic Energy

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Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley Goals for Chapter 6 To understand and calculate work done by a force To study and apply kinetic energy To learn and use the work-energy theorem To see the difference from the first bullet and then calculate work done by a varying force along a curved path To add time to the calculation and determine the power in a physical situation

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Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley Introduction We’ve studied how Newton’s Second Law allows us to calculate an acceleration from a force but what if the force changes during its application? We must be able to account for things like an archer’s bow. We must look at action– reaction pairs that are not immediately obvious (like the shotgun expelling the pellets with expanding gas but having the expanding gas do work on the shotgun at the same

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Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley 6.1 Work Caution: Work = W, weight = w Work is done when a force exerted on a body caused the body to undergo a displacement. Work W is defined as a dot product: The SI unit of work is the joule 1 joule = (1 newton)·(1 meter) or 1 J = 1 N∙m 1 J = 1 kg∙m 2 /s 2

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Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley Work is a scalar quantity Even though work is calculated by using two vector quantities (force and displacement), it is a dot product. Dot product is a scalar quantity. It has only a number value. It has no direction! F∙s = F x s x + F y s y + F z s z

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Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley Example 6.1 - Use the parallel component if the force acts at an angle a.Steve exerts a steady force of magnitude 210 N on the stalled car as he pushes it a distance of 18 m. The car also has a flat tire, so to make the car track straight Steve must push at and angle of 30 o to the direction of motion. How much work does Steve do? b. in a helpful mood, Steve pushes a second stalled car with a steady force The displacement of the car is How much work does Steve do in this case? W = 3300 J W = 1800 J

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Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley Positive Work, Negative Work and No Work When 0 ≤ Φ < 90 o work is positive When 90 o < Φ ≤ 180 o work is negative When Φ = 90 o work is zero

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Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley How can it be such a great “workout” with no work? When positive and negative work cancel, the net work is zero even though muscles are exercising. CAUTION keep track of who’s doing the work The total work W tot done on the body is the algebraic sum of the quantities of work done by the individual forces.

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Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley Example 6.2- solution of work done by several forces A farmer hitches her tractor to a sled loaded with firewood and pulls it a distance of 20 m along level ground. The total weight of sled and load is 14,700 N. the tractor exerts a constant 5000-N force at an angle of 36.9o above the horizontal. There is a 3500 N friction force opposing the sled’s motion. Find the work done by each force acting on the sled and the total work done by all the forces. W T = 80 kJ W f = -70 kJ W tot = 10 kJ F x = 500 N W tot = (500)(20m)=10 kJ

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Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley Test Your Understanding 6.1 An electron moves in a straight line toward the east with a constant speed of 8 x 10 7 m/s. It has electric, magnetic, and gravitational forces acting on it. During a 1-m displacement, the total work done on the electron is 1.Positive 2.Negative 3.Zero 4.Not enough information given to decide. Answer: iii Since the electron has constant velocity, its acceleration is zero, and the net force on the electron is also zero. Therefore the total work done by all the forces must be zero as well.

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Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley 6.2 kinetic energy and work-energy theorem The work done by the net force on a particle equals the change in the particle’s kinetic energy:

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Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley We can compare the kinetic energy of different bodies Kinetic energy and work have the same units: 1 J = 1 N∙m = 1 kg∙m 2 /s 2

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Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley Example 6.3 Let’s look again at Example 6.2. Suppose the initial speed v 1 is 2.0 m/s. What is the speed of the sled after it moves 20 m? v 2 = 4.2 m/s

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Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley Example 6.4 - Forces on a hammerhead In a pile driver, a steel hammerhead with mass 200 kg is lifted 3.00 m above the top of a vertical I-beam being driven into the ground. The hammer is then dropped, driving the I-beam 7.4 cm farther into the ground. The vertical rails that guide the hammerhead exert a constant 50 N friction force on the hammerhead. Use the work-energy theorem to find a.The speed of the hammerhead just as it hits the I-beam b.The average force the hammerhead exerts on the I-beam. Ignore the effects of the air. a. v = 7.55 m/s b. F = 78900 N

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Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley Different objects, different kinetic energies Two iceboats hold a race on a frictionless horizontal lake. The two iceboats have masses m and 2m. Each iceboat has an identical sail, so the wind exerts the same constant force F on each iceboat. The two iceboats start from rest and cross the finish line a distance s away. Which iceboat crosses the finish line with greater kinetic energy? same

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Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley Test Your Understanding 6.2 Rank the following bodies in order of their kinetic energy, from least to greatest. i.A 20.0 kg body moving at 5.0 m/s ii.A 1.0 kg body that initially was at rest and then had 30 J of work don on it; iii.A 1.0 kg body that initially was moving at 4.0 m/s and then had 30 J of work done on it; iv.A 20.0 kg body that initially was moving at 10 m/s and then did 80 J of work on another body. iv > i > iii > ii

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Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley 6.3 Work and energy with varying forces Perhaps the best example is driving a car, alternating your attention between the gas and the brake. The effect is a variable positive or negative force of various magnitude along a straight line. Area = work

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The integral represents the area under the curve between x 1 and x 2. On a graph of force as a function of position, the total work done by the force is represented by the area under the curve between the initial and final positions.

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Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley The stretch of a spring and the force that caused it The force applied to an ideal spring will be proportional to its stretch. The graph of force on the y axis versus stretch on the x axis will yield a slope of k, the spring constant.

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When spring is compressed, Hooke’s law still holds. Both force and displacement are negative, so the total work is still positive.

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Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley Example 6.6 Stepping on a scale A woman weighing 600 N steps on a bathroom scale containing a stiff spring. In equilibrium the spring is compressed 1.0 cm under her weight. Find the force constant of the spring and the total work done on it during the compression., k = 6.0x10 4 N/m W = 3.0 J

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Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley The work-energy theorem, W tot = K 2 – K 1, is true even when the force varies with position. Work-Energy Theorem for Straight-Line Motion, Varying Forces

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Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley Example 6.7 Motion with a varying force An air-track glider of mass 0.100 kg is attached to the end of a horizontal air track by a spring with force constant 20.0 N/m. Initially the spring is unstretched and the glider is moving at 1.50 m/s to the right. Find the maximum distance d that the glider moves to the right a)If the air track is turned on so that there is no friction. b)If the air is turned off so that there is kinetic friction with coefficient µ k = 0.47 a. 10.6 cm b. 8.6 cm

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Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley Work-Energy Theorem for Motion Along a Curve W tot = ∆K = K 2 – K 1 is true in general, no matter what the path and no matter what the character of the forces.

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Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley Example 6.8 Motion on a curved path At a family picnic you are appointed to push your cousin Joe in a swing. His weight is w, the length of the chains is R, and you push Joe until the chains make an angle θ o with the vertical. To do this, you exert a varying horizontal force F that starts at zero and gradually increases just enough so that Joe and the swing move very slowly and remain very nearly in equilibrium. a.What is the total work done on Joe by all forces? b.What is the work done by the tension T in the chains? c.What is the work you do by exerting the force F? a.0 b.0 c.wR(1-cosθ o )

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Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley Test Your Understanding 6.3 In a conical pendulum, if the speed of the penulum bob remains constant as it travels around the circle, a.Over one complete circle, how much work doens the tension force F do on the bob? i.A positive amount ii.A negative amount iii.Zero b.Over once complete circle, how much work does the weight do on the bob? i.A positive amount ii.A negative amount iii.Zero. a.0 (iii) b.0 (iii)

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Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley 6.4 power Power is the time rate at which work is done. Like work and energy, power is a scalar quantity.

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Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley 6.4 Watt about power? Once work is calculated, dividing by the time that passed determines power. The pun is credit to James Watt, the inventor of first modern steam engine. Another power unit – horsepower The energy you use may be noted from the meter the electric company probably installed to measure your consumption of energy in kilowatt-hours. 1 hp = 746 watts

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In mechanics we can also express power in terms of force and velocity. Suppose that a force F acts on a body while it undergoes a vector displacement s. If F // is the component of F tangent to the path (parallel to ∆s), then the work done by the force is ∆W = F // ∆s. the average power is Instantaneous power P is the limit of this expression as t approaches to zero

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Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley Example 6.10 Each of the two jet engines in a Boeing 767 airliner develops a thrust (a forward force) of 197,000 N. When the airplane is flying at 250 m/s, what horsepower does each engine develop? 66.000 hp

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Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley Example 6.11 A 50.0 kg marathon runner runs up the stairs to the top of Chicago’s 443 m tall Sears Tower, the tallest building the United States. To lift herself to the top in 15.0 minutes, what must be her average power output in watts? In kilowatts? In horsepower? 241 W 0.241 kW 0.323 hp

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