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Physics 207, Lecture 10, Oct. 9 • • MidTerm I Exams will be returned in your next discussion section Regrades: Write down, on a separate sheet, what you want regraded and why. Mean: 64.6 Median: 67 Std. Dev.: 19.0 Range: High 100 Low 5 Solution posted on http://my.wisc.edu Nominal curve (conservative): 86-100 A 70-85 B or A/B 40-69 C or B/C 35-40 marginal 25-35 D Below 25 F Physics 207: Lecture 10, Pg 1 Physics 207, Lecture 10, Oct. 9 Agenda: Chapter 7, Work and Energy Transfer • Definition of Work (a scalar quantity) • Variable force devices (e.g., Hooke’s Law spring) • Work/Energy Theorem • • W = DK Kinetic Energy K = 1/2 mv2 Power P = dW / dt = F · v Assignment: For Wednesday read Chapter 8 WebAssign Problem Set 4 due Tuesday next week (start now) Physics 207: Lecture 10, Pg 2 Work & Energy One of the most important concepts in physics. Alternative approach to mechanics. Many applications beyond mechanics. Thermodynamics (movement of heat or particles). Quantum mechanics... Very useful tools. You will learn a complementary approach (often much easier) way to solve problems. But there is no free lunch….easier but there are fewer details that are explicitly known. Physics 207: Lecture 10, Pg 3 See text: 7-1 Energy Conservation Energy cannot be destroyed or created. Just changed from one form to another. We say energy is conserved ! True for any isolated system. Doing “work” on an otherwise isolated system will change it’s “energy”... Physics 207: Lecture 10, Pg 4 See text: 7-1 Definition of Work, The basics Ingredients: Force ( F ), displacement ( D r ) Work, W, of a constant force F acting through a displacement D r is: W = F ·D r F Dr (Work is a scalar) “Scalar or Dot Product” Work tells you something about what happened on the path! Did something do work on you? Did you do work on something? Simplest case (no frictional forces and no non-contact forces) Did your speed change? ( what happened to | v | !!!) Physics 207: Lecture 10, Pg 5 Remember that a path evolves with time and acceleration implies a force acting on an object v path and time a t a = atang + aradial a a a = a + a a =0 F = Ftang + Fradial Two possible options: Change in the magnitude of v a =0 Change in the direction of v a =0 A tangetial force is the important one for work! How long (time dependence) gives the kinematics The distance over which this forceTang is applied: Work Physics 207: Lecture 10, Pg 6 Definition of Work... Only the component of F along the path (i.e. “displacement”) does work. The vector dot product does that automatically. Example: Train on a track. F Dr F cos If we know the angle the force makes with the track, the dot product gives us F cos and Dr Physics 207: Lecture 10, Pg 7 Review: Scalar Product (or Dot Product) 7.3 Useful for performing projections. A A î = Ax î Ay Ax Calculation is simple in terms of components. A B = (Ax )(Bx) + (Ay )(By ) + (Az )(Bz ) Calculation also in terms of magnitudes and relative angles. A B ≡ | A | | B | cos You choose the way that works best for you! Physics 207: Lecture 10, Pg 8 See text: 7-1 Work: 1-D Example (constant force) A force F = 10 N pushes a box across a frictionless floor for a distance Dx = 5 m. Finish Start F = 0° Dx Work is A B ≡ | A | | B | cos = F Dx = 10 x 5 N m = 50 J 1 Nm is defined to be 1 Joule and this is a unit of energy Work reflects energy transfer See example 7-1: Pushing a trunk. Physics 207: Lecture 10, Pg 9 See text: 7-1 Units: Force x Distance = Work Newton x Meter = Joule [M][L] / [T]2 [L] [M][L]2 / [T]2 mks N-m (Joule) cgs Dyne-cm (erg) = 10-7 J Other BTU calorie foot-lb eV = 1054 J = 4.184 J = 1.356 J = 1.6x10-19 J Physics 207: Lecture 10, Pg 10 See text: 7-1 Work: 1-D 2nd Example (constant force) A force F = 10 N pushes a box across a frictionless floor for a distance Dx = 5 m. Finish Start = 180° F Dx Work is A B ≡ | A | | B | cos = F Dx (-1) = -10 x 5 N m = -50 J Work reflects energy transfer See example 7-1: Pushing a trunk. Physics 207: Lecture 10, Pg 11 See text: 7-1 Work: 1-D 3rd Example (constant force) A force F = 10 N pushes a box across a frictionless floor for a distance Dx = 5 m. Start F Finish = -45° Dx Work is A B ≡ | A | | B | cos = F Dx 0.71= 50 x 0.71 Nm = 35 J Work reflects energy transfer See example 7-1: Pushing a trunk. Physics 207: Lecture 10, Pg 12 Text : 7.3 Work and Varying Forces Area = Fx Dx F is increasing Here W = F ·D r becomes dW = F dx xf Consider a varying force F(x) Fx W x Dx xi Finish Start F F ( x ) dx F = 0° Dx Work is a scalar, the catch is that there is no time/position info on hand Physics 207: Lecture 10, Pg 13 Lecture 10, Exercise 1 Work in the presence of friction and non-contact forces A box is pulled up a rough (m > 0) incline by a ropepulley-weight arrangement as shown below. How many forces are doing work on the box ? Of these which are positive and which are negative? Use a Force Body Diagram (A) 2 Compare force and path (B) 3 v (C) 4 Physics 207: Lecture 10, Pg 14 Lecture 10, Exercise 1 Work in the presence of friction and non-contact forces A box is pulled up a rough (m > 0) incline by a ropepulley-weight arrangement as shown below. How many forces are doing work on the box ? And which are positive and which are negative? Use a Force Body Diagram (A) 2 v T (B) 3 is correct N (C) 4 mg f Physics 207: Lecture 10, Pg 15 Text : 7.3 A variable force device: A Hooke’s Law Spring Springs are everywhere, (probe microscopes, DNA, an effective interaction between atoms) Rest or equilibrium position Dx F(work done on spring) In this spring, the magnitude of the force increases as the spring is further compressed (a displacement). Hooke’s Law, Active Figure FS = - k Dx Dx is the amount the spring is stretched or compressed from it resting position. Physics 207: Lecture 10, Pg 16 Lecture 10, Exercise 2 Hooke’s Law Remember Hooke’s Law, Fx = -k Dx What are the units for the constant k ? (A) kg m 2 s2 (B) kg m s2 (C) kg s2 (D) kg 2 m s2 F is in kg m/s2 and dividing by m gives kg/s2 or N/m Physics 207: Lecture 10, Pg 17 Lecture 10, Exercise 3 Hooke’s Law 8 cm 9 cm What is the spring constant “k” ? 0.50 kg (A) 50 N/m (B) 100 N/m (C) 400 N/m (D) 500 N/m Physics 207: Lecture 10, Pg 18 Lecture 10, Exercise 3 Hooke’s Law 8 cm 9 cm What is the spring constant “k” ? Fspring 0.50 kg (A) 50 N/m SF = 0 = Fs – mg = k Dx - mg Use k = mg/Dx = 5 N / 0.01 m (B) 100 N/m (C) 400 N/m (D) 500 N/m mg Physics 207: Lecture 10, Pg 19 Force (N) F-x relation for a foot arch: Displacement (mm) Physics 207: Lecture 10, Pg 20 F-x relation for a single DNA molecule Physics 207: Lecture 10, Pg 21 Measurement technique: optical tweezers Physics 207: Lecture 10, Pg 22 Work & Kinetic Energy: Energy transfer involving changes in speed A force, F = 10 N, pushes a box across a frictionless floor for a distance Dx = 5m. The speed of the box is v1 before the push, and v2 after the push. Consider only this force and the box Relate the work to the kinetic energy of the box F v v1 F’ 2 m i Dx Physics 207: Lecture 10, Pg 23 See text: 7-4 Work Kinetic-Energy Theorem: {Net Work done on object} = {change in kinetic energy of object} Wnet DK K 2 K1 (final – initial) 1 1 2 2 mv 2 mv1 2 2 Physics 207: Lecture 10, Pg 24 Example: Work Kinetic-Energy Theorem • How much will the spring compress (i.e. Dx) to bring the object to a stop (i.e., v = 0 ) if the object is moving initially at a constant velocity (vo) on frictionless surface as shown below ? to vo Notice that the spring force is opposite to the displacemant. F m spring at an equilibrium position For the mass m, work is negative Dx V=0 t m For the spring, work is positive spring compressed Physics 207: Lecture 10, Pg 25 Example: Work Kinetic-Energy Theorem • How much will the spring compress (i.e. Dx = xf - xi) to bring the object to a stop (i.e., v = 0 ) if the object is moving initially at a constant velocity (vo) on frictionless surface as shown below ? xf to vo Wbox F xi xf m spring at an equilibrium position Dx V=0 t F ( x ) dx Wbox xi x f Wbox - kx | xi 1 2 Wbox m spring compressed kx dx 2 - k Dx DK 1 2 2 - k Dx m0 mv 1 2 2 1 2 2 1 2 2 0 Physics 207: Lecture 10, Pg 26 Lecture 10, Exercise 4 Kinetic Energy To practice your pitching you use two baseballs. The first time you throw a slow curve and clock the speed at 50 mph (~25 m/s). The second time you go with high heat and the radar gun clocks the pitch at 100 mph. What is the ratio of the kinetic energy of the fast ball versus the curve ball ? (A) 1/4 (B) 1/2 (C) 1 (D) 2 (E) 4 Physics 207: Lecture 10, Pg 27 Lecture 10, Exercise 4 Kinetic Energy To practice your pitching you use two baseballs. The first time you throw a slow curve and clock the speed at 50 mph (~25 m/s). The second time you go with high heat and the radar gun clocks the pitch at 100 mph. What is the ratio of the kinetic energy of the fast ball versus the curve ball ? KE2/KE1 = ½ mv22 / ½ mv12 = 1002 / 502 = 4 (A) 1/4 (B) 1/2 (C) 1 (D) 2 (E) 4 Physics 207: Lecture 10, Pg 28 Lecture 10, Exercise 5 Work & Friction Two blocks having mass m1 and m2 where m1 > m2. They are sliding on a frictionless floor and have the same kinetic energy when they encounter a long rough stretch (i.e. m > 0) which slows them down to a stop. Which one will go farther before stopping? Hint: How much work does friction do on each block ? (A) m1 (B) m2 m1 m2 (C) They will go the same distance v1 v2 Physics 207: Lecture 10, Pg 29 Lecture 10, Exercise 5 Work & Friction W = F d = - m N d = - m mg d = DK = 0 – ½ mv2 - m m1g d1 = - m m2g d2 d1 / d2 = m2 / m1 (A) m1 (B) m2 m1 m2 (C) They will go the same distance v1 v2 Physics 207: Lecture 10, Pg 30 Lecture 10, Exercise 6 Work & Friction You like to drive home fast, slam on your brakes at the start of the driveway, and screech to a stop “laying rubber” all the way. It’s particularly fun when your mother is in the car with you. You practice this trick driving at 20 mph and with some groceries in your car with the same mass as your mother. You find that you only travel half way up the driveway. Thus when your mom joins you in the car, you try it driving twice as fast. How far will you go this time ? A. The same distance. Not so exciting. B. 2 times as far (only ~7/10 of the way up the driveway) C. Twice as far, right to the door. Whoopee! D. Four times as far crashing into the house. (Oops.) Physics 207: Lecture 10, Pg 31 Lecture 10, Exercise 6 Work & Friction W = F d = - m N d = - m mg d = DK = 0 – ½ mv2 W1= - m mg d1= DK1= 0 – ½ mv12 W2= - m mg d2= DK2= 0 – ½ m(2v1)2 = – 4 (½ mv12) - m mg d2= – 4 (m mg d1) d2= – 4 d1 A. The same distance. Not so exciting. B. 2 times as far (only ~7/10 of the way up the driveway) C. Twice as far, right to the door. Whoopee! D. Four times as far crashing into the house. (Oops.) Physics 207: Lecture 10, Pg 32 Work & Power: Two cars go up a hill, a Corvette and a ordinary Chevy Malibu. Both cars have the same mass. Assuming identical friction, both engines do the same amount of work to get up the hill. Are the cars essentially the same ? NO. The Corvette gets up the hill quicker It has a more powerful engine. Physics 207: Lecture 10, Pg 33 Work & Power: Power is the rate at which work is done. W Average Power is, P Instantaneous Power is, Dt dW P dt Physics 207: Lecture 10, Pg 34 Lecture 10, Exercise 8 Work & Power Starting from rest, a car drives up a hill at constant acceleration and then suddenly stops at the top. The instantaneous power delivered by the engine during this drive looks like which of the following, (A) (B) (C) time time time Physics 207: Lecture 10, Pg 35 Lecture 10, Exercise 8 Work & Power P = dW / dt and W = F d = (m mg cos mg sin ) d and d = ½ a t2 (constant accelation) So W = F ½ a t2 P = F a t = F v (A) time (B) time (C) time Physics 207: Lecture 10, Pg 36 Work & Power: Power is the rate at which work is done. Average Power: W P Dt Instantaneous Power: Units (SI) are Watts (W): dW 1 W = 1 J / 1s P dt Simple Example 1 : A person of mass 80.0 kg walks up to 3rd floor (12.0m). If he/she climbs in 20.0 sec what is the average power used. Pavg = F h / t = mgh / t = 80.0 x 9.80 x 12.0 / 20.0 W P = 470. W Physics 207: Lecture 10, Pg 37 Work & Power: Simple Example 2 : Engine of a jet develops a trust of 15,000 N when plane is flying at 300 m/s. What is the horsepower of the engine ? dW P dt W F Dx dW F dx dW dx P F dt dt P F v P=Fv P = (15,000 N) (300 m/s) = 4.5 x 106 W = (4.5 x 106 W) (1 hp / 746 W) ~ 6,000 hp ! Physics 207: Lecture 10, Pg 38 Physics 207, Lecture 10, Recap Agenda: Chapter 7, Work and Energy Transfer • Definition of Work (a scalar quantity) • Variable force devices (e.g., Hooke’s Law spring) • Work/Energy Theorem • • W = DK Kinetic Energy K = 1/2 mv2 Power P = dW / dt = F · v Assignment: For Wednesday read Chapter 8 WebAssign Problem Set 4 due Tuesday next week (start now) Physics 207: Lecture 10, Pg 39