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What causes motion In our everyday life we observe that objects often change their state of motion. A lot that happens in the Universe results from a change in motion. These changes are produced by forces and in our everyday life there are just two forces we can sense directly. Gravity acts on mass F = Gm1m2/r2 F = kq1q2/r2 Electromagnetic forces act on electric charge Week 3, Physics 214 Spring 2010 1 FORCES Forces also bind objects together, and can even change the composition of mattter – for example the electromagnetic force is responsible for all of chemistry and for the structure of atoms. Electromagnetic forces cause opposite electric charges to attract each other, as in the hydrogen atom where the positive proton (which is the nucleus) attracts the orbiting electron and binds it into the atom. F = kq1q2/r2 Week 3, Physics 214 Spring 2010 2 Newtons Second and First Law Second Law The acceleration of an object is directly proportional to the magnitude of the imposed force and inversely proportional to the mass. The acceleration is in the same direction as the force F = ma F and a are vectors unit is a Newton (or pound) 1lb = 4.448N First Law An object remains at rest or in uniform motion in a straight line unless it is acted on by an external force. F = 0 a = 0 so v = constant http://www.physics.purdue.edu/academic_programs/courses/phys214/movies.php (anim0003.mov) (anim0004.mov) Week 3, Physics 214 Spring 2010 3 Newtons Second and First Law Second Law F = ma F and a are vectors First Law is just a special case of the Second Law An object remains at rest or in uniform motion in a straight line unless it is acted on by an external force. F = 0 : a = 0 so v = constant NOTE: if a = 0, you can conclude that the NET force on the object is ZERO. The NET force is the vector sum of ALL forces acting on the object. http://www.physics.purdue.edu/academic_programs/courses/phys214/movies.php (anim0003.mov) (anim0004.mov) Week 3, Physics 214 Spring 2010 4 Ch 4 E6 & quiz A 6kg block is being pushed with a force P and has an acceleration of 3.0m/s2 a) What is the net force? b) If P is 20N what is Ff? Note red arrows are NOT in proportion!! P Ff A 3N, B 6N , a) F = ma = 6 x 3 = 18 N b) F = P – Ff Week 3, a = 3m/s2 6 kg D 18 N , C 12 N E 24 N Ff = 2N Physics 214 Spring 2010 5 Mass and weight Newtons second law enables us to measure relative mass. If we apply the same force to two objects and measure the accelerations then. F = m1a1 and F = m2a2 so m1/m2 = a2/a1 We then need to have one mass as a calibration. A kilogram is defined as the mass of a special piece of platinum held in Paris. Measuring acceleration is a more difficult way of measuring mass than using a spring or a balance. The balance exploits the fact that gravity acts proportional to mass, so that at a given place all masses accelerate equally. Week 3, Physics 214 Spring 2010 6 Mass and weight Since gravity acts proportional to mass (this is Newton’s law of Universal Gravitation) then the force near the earths surface is F = mg This force is known as the weight of an object so if we compare F1 = m1g and F2 = m2g then weight1/weight2 = m1/m2 An equal-arm balance has equal weights on each side when it is “balanced”. Unequal-arms use the fact of leverage to measure relative weights. Week 3, Physics 214 Spring 2010 7 Mass and weight A spring exerts a force proportional to the force pulling it. So a mass hung on a spring will stretch the spring proportional to its weight, W, which is proportional to its mass: W = mg This method is cheap and quick and typically less accurate than comparing two masses with a balance. m mg Week 3, Physics 214 Spring 2010 8 Inertia Inertia = tendency of an object to resist changes in its velocity. Since F = ma and a = Δv/t then Ft = mΔv : impulse So if a force acts for a time INTERVAL t the change in velocity will be smaller for larger masses. So it is mass that determines inertia. In particular if t is very small and m is large, then F can be large while Δv is very small. NOTE: for a given impulse (mΔv) if t (actually, Δt) is really small, then F must be very large. PADDING!: Week 3, Physics 214 Spring 2010 9 Friction In our everyday world any object which is moving feels a force opposing the motion --- this is friction. An object which is sliding The air resistance on your car These types of friction result in energy being lost. So minimizing friction is very important. But Friction is also useful and essential since with no friction a car would not move but just spin it’s wheels a car would not be able to turn a corner we would not be able to walk objects would slide off surfaces unless perfectly horizontal Week 3, Physics 214 Spring 2010 10 Force diagrams When we are analyzing a particular object we have to take into account ALL the forces acting on the body both in magnitude and direction. (i.e. vector sum of all Forces) The acceleration of the object is equal to F/m in the direction of F where F is the NET force acting. As in the example above we know that, if we want to move an object --- there is a force called friction which opposes what we want to do. In the case shown the 10N force is + and the 2N force (-) so the net force is 8N and 8N = 5kg x a so a = 1.6[N/kg or m/s2] Check our work: a Week 3, has units m/s/s and N = kg m/s/s Looks OK Physics 214 Spring 2010 11 Friction In the case shown the 10N force is + and the 2N force - so the net force is 8N So: 8N = 5kg x a There can be sliding friction and there can be static (“stuck”) friction. Stuck (static) friction opposes other forces, UP TO A LIMIT. Beyond which it “breaks”. In the present example, since the NET force is not zero, the object must be accelerating and the friction must be sliding friction. There are cases where the other forces aren’t big enough to break the static friction: then the static friction adjusts itself to just oppose the other forces, and the NET force is zero. Hence no acceleration and no motion. The maximum static friction is proportional to the size of the force pushing the object towards the surface, and to the “coefficient of friction” : small c.o.f. = slick, big c.o.f. = good traction. Week 3, Physics 214 Spring 2010 12 Friction T f = μk(W - Tvertical) Thorizontal W = mg The maximum static friction is proportional to the size of the force pushing the object towards the surface, and to the “coefficient of friction” : small c.o.f. = slick, big c.o.f. = good traction. a = (Thorizontal – f )/m It’s better to pull upwards than to push downwards on a box or a suitcase you’re trying to slide sideways !! Week 3, Physics 214 Spring 2010 13 Terminal velocity As an object moves through the atmosphere the air exerts a frictional retarding force (drag) which increases with velocity. An object that is dropped from a great height first accelerates at 9.8m/s2. As the speed increases, the drag increases and cancels more and more of the gravity force. At TERMINAL velocity, the drag force just balances the weight (gravity) force: NET force is zero, acceleration is zero, so the velocity does not change any further. Hence “terminal” velocity. Ff mg v Mg – Ff =ma Similarly when a car is moving at constant speed on the highway the forward force of the road on the tires is balanced by the frictional force of the air on the car. Fair Ftire Freaction – Fair = ma Freaction Week 3, Physics 214 Spring 2010 14 Re-action : Newtons Third Law For every force that acts on a body there is an equal and opposite reaction force (by the body ON the agent.) block pulls you –F NET on block you pull block F F F f floor on block * * net -Ff block on floor You pull on the block; the block pulls back on you The floor exerts a frictional force holding the block back The block exerts a frictional force on the floor trying to move the floor to the right. To find effect on block, use ALL, and ONLY, Forces* acting ON the block The block accelerates providing F > Ff i.e. Fnet >0 Week 3, Physics 214 Spring 2010 15 Force analysis & quiz To analyze the motion of an object we need to draw a diagram and put in all the forces that are acting ON the object. We will only deal with problems that have an acceleration along a single axis. The net force along an axis perpendicular to this axis is zero A B Ff Fnet D F -Ff C We can choose + in the direction of the acceleration. We now can use the equations v = v0 + at d = v0t + ½ at2 v2 = v02 + 2ad Fnet = ma Which one of the above force vectors did we ignore? Week 3, Physics 214 Spring 2010 16 Reaction forces IN THESE DIAGRAMS, THE NET FORCES ON OBJECTS ARE ZERO N For the book, magnitudes Nt = mg Nt, table on book, is vector pointing opposite to mg For the table, magnitude Nearth = mg + mtableg mg mtableg mg Ne Here the situation is static, that is, no acceleration (and no motion). The “Normal” forces are support forces, and ADJUST themselves to prevent motion – up to the point where something breaks. Static friction is a little bit like this too – it adjusts itself up to the point where the object breaks free and starts to move. Week 3, Physics 214 Spring 2010 17 Reaction forces N For the book, magnitudes Nt = mg Nt, table on book, is vector pointing opposite to mg For the table, magnitudes Nearth = mg + mtableg mg mtableg mg Ne For a stack of plates or a brick wall and even a mountain each layer has to support the weight of everything higher. So for a stack of 48 plates the force on the second plate from the top is mg. For the bottom plate it is 47mg. Of course each plate has a NET Normal force of +mg to balance it’s own weight, -mg. Week 3, Physics 214 Spring 2010 18 I-clicker Registration Would the following students please register your iclickers in CHIP so that your answers to the Lecture Quizzes can be entered? Bic Blackwell Chan O’Keefe Walter Week 3, Physics 214 Spring 2010 19 Forces in an accelerating elevator W = mg = true weight with no acceleration + N = apparent weight = “scale on object” N N – mg is the net force, where + means up N – mg = ma is the equation of motion g If N > mg a is positive and the apparent weight is > than the true weight If N < mg a is negative and the apparent weight is less than the true weight mg If N < 0, scale & floor aren’t supporting and object is in free fall at 1 g of acceleration downward IRRESPECTIVE OF THE DIRECTION OR MAGNITUDE OF THE VELOCITY Week 3, Physics 214 Spring 2010 20 Connected objects T Both objects have the same acceleration The coupling pulls back on the 4kg mass but accelerates the 2kg mass. 30N – 8N – T = 4kg a T – 6N = 2kg a Two equations, and two unknowns (T and a). You can eliminate T between the two eqns., then solve for a. (Just add the 2 eqns. This is equivalent to treating the two blocks as one system of mass 6 kg, with total friction 14 N, and a common acceleration.) 30N – 8N – 6N = 6kg a ; a = 16/6 m/s2 ; T = 6N+ 16/3N T = 111/3 N [N/kg=m/s2] Week 3, Physics 214 Spring 2010 21 Connected objects If we have a freight train with 100 cars each of mass m, and the train is accelerating at a rate a , then for the coupling between the engine and the first car T1 = 100ma And between the last two cars T99 = ma or 100 times less since that tension has to accelerate only ONE car. Week 3, Physics 214 Spring 2010 22 Motion of a car Fair Ftire Freaction – Fair = ma Freaction Generally Fair is proportional to v2 . And energy lost to air drag = Force x Distance. So the difference in energy lost to air resistance between 55mph and 80mph is a factor of 2.16. Air resistance is however only part of the losses, which include friction in the drive train, friction in the wheel bearings, and friction in the tires as they flex, among other things. Assuming these other frictional losses ALSO go as v2, (which may be an overstatement) you use (80/55)2 = 2.16 more gasoline to cover the same distance. Travel from Indianapolis to Lafayette a distance of 60 miles Car does 30miles/gallon at 55mph gas costs $3/gallon At 55mph use 60mi/(30mi/gal) = 2 gallons cost $6 time = 60mi/(55mi/hr) = 1.0909 hr = 65.45 minutes At 80 mph use 4.32 gallons cost $12.96 time = 60mi/(80mi/hr) = 0.75 hr = 45 minutes Cost of saving 20.45 minutes = $6.96 Week 3, Physics 214 Spring 2010 23 1H-03 CO2 Rocket Imagine that you are sitting in a cart with a pile of bricks. ROCKET PROPULSION ! How could you use the bricks to get yourself and the cart to move ? What would happen if you throw a brick out of the cart ? Then you throw out another . What if you throw smaller bricks faster and more frequently ? Now, if the bricks were the size of molecules. . . What happens when the fire extinguisher rapidly “throws” out CO 2 molecules ? THIS IS DIFFERENT THAN THE FAN CART. CO2 IS EXPELLED AT HIGH VELOCITY AND IN Because the repulse TERMS OF FORCES THE REACTION FORCE CAUSES THE CART TO MOVE IN THE OPPOSITE DIRECTION. THE QUANTITY CALLED MOMENTUM IS CONSERVED AND THE MASS OF THE CO2 X AVERAGE SPEED = TOTAL MASS OF THE CART X AVERAGE VELOCITY. THIS IS HOW A ROCKET IS ACCELERATED AND IT ALSO WORKS IN OUTER SPACE. Week 3, Physics 214 Spring 2010 24 Summary of Chapter 4 Forces are responsible for all physical phenomena Gravitation and the electromagnetic force are responsible for all the phenomena we normally observe in our everyday life. Newton’s laws F = ma where F is net force on m d = v0t + ½ at2 d = ½(v + v0)t v2 = v02 + 2ad v = v0 + at Every force produces an equal and opposite reaction Fa on b = -Fb on a Weight = mg where g = 9.8m/s2 locally Apparent “weight” in an elevator depends on the acceleration a up : “weight” is higher lower a down: “weight” is If your “weight” becomes zero it’s time to worry because you are in free fall!! Week 3, Physics 214 Spring 2010 25 Ch 4 E16 g N A vertical force of 6N presses on a book. a) What is the gravitational force? b) What is the normal force on the book? 6N 0.4 kg a) Gravitational Force = mg = 3.92 N b) Upward Force = 6 + 3.92 = 9.92 N Week 3, Physics 214 Spring 2010 26 1C-04 Bocce Ball Tracks Two balls are released with the same initial velocity. One travels a flat path the other goes down a slope increasing it’s speed and then climbs a slope and slows down before reaching the end. The balls roll so friction is small. V V V Which ball will reach the end first ? 1 5 2 3 4 The time for the yellow ball is just d/v. The time for the blue ball can be computed in 5 parts. Parts 1 and 5 are identical for both balls but the blue ball is faster for part 3 and qualitatively one would expect the blue ball to arrive first. Week 3, Physics 214 Spring 2010 27 1F-04 Brass Rod (Inertia) How to remove the paper without toppling the rod ? One needs to remove the paper quickly so that the frictional force only lasts for a short time and the inertia of the rod prevents it from toppling over. Week 3, Physics 214 Spring 2010 28 1F-05 Coin, Hoop & Milk Bottle (Inertia) How can you get the coin into the bottle without touching it ? This is actually a trick which depends on hitting the ring so that the top deflects down and the coin is free to drop Week 3, Physics 214 Spring 2010 29 1F-06 Inertial Ball Which string breaks first ? Case 1: Place the aluminum rod in the lower loop and pull SLOWLY downward. Case 2: Use the wooden mallet to strike a sharp blow to the aluminum rod. IF IT IS DONE SLOWLY THE UPPER STRING BREAKS FIRST BECAUSE THE TENSION IN THAT STRING WILL BE THE WEIGHT OF THE BALL PLUS THE TENSION IN THE LOWER STRING. IF THE LOWER STRING IS STRETCHED SUFFICIENTLY RAPIDLY, IT WILL REACH ITS BREAKING POINT BEFORE THE BALL HAS A CHANCE TO MOVE APPRECIABLY. Week 3, Physics 214 Spring 2010 30 1F-02 Stack of Washers This is a demonstration of Inertia where a washer can be removed from a stack if the blow is fast. Why does it work less well as the stack gets shorter ? Strike the stack quickly. So the friction will be very short-lived and the stack will not gain speed before the force is gone. THIS TENDENCY TO RESIST CHANGES IN THEIR STATE OF MOTION IS DESCRIBED AS INERTIA. MASS IS A MEASURE OF THE AMOUNT OF INERTIA. SO FOR A FIXED FRICTIONAL FORCE ACTING FOR A SHORT TIME. THE BIGGER THE MASS, THE LESS IT WILL MOVE. Week 3, Physics 214 Spring 2010 31 1F-07 Table Cloth Jerk Can the table cloth be removed without breaking any dishes ? THERE IS A FORCE ACTING ON THE DISHES, BUT IT LASTS FOR A VERY SHORT TIME. COMBINED WITH THE RELATIVELY LARGE MASS OF THE DISHES, THIS FORCE IS OVER SO QUICKLY AND IS SO SMALL THAT THE DISHES HARDLY MOVE. Week 3, Physics 214 Spring 2010 32 1F-03 Egg drop Is it possible to get the eggs in the beakers without touching them ? IF THE PAN IS HIT SHARPLY A FORCE WILL ACT ON THE EGGS FOR A VERY SHORT TIME AND THEY WILL NOT MOVE HORIZONTALLY. THE PAN HAS TO BE HIT HARD ENOUGH SO THAT HAS MOVED OUT OF THE WAY BEFORE THE EGGS DROP ANY APPRECIABLE DISTANCE Week 3, Physics 214 Spring 2010 33 1H-02 Fan Cart (Action-Reaction) Can a fan attached to a cart propel the cart? What if the sail is removed? Reaction Action In which direction will it move ? What if the sail is canted at an angle ? INTERNAL FORCES IN A SYSTEM CANCEL EACH OTHER WHEN THE SYSTEM AS A WHOLE IS CONSIDERED. SO IF THE SAIL IS PERPENDICULAR THE FAN DOES NOT MOVE. IF THE SAIL IS REMOVED THE FAN MOVES IN THE OPPOSITE DIRECTION TO WHICH IT BLOWS AIR. THE FAN WOULD NOT MOVE IN OUTER SPACE. Week 3, Physics 214 Spring 2010 34 1H-04 Hero's Engine A glass bulb emits steam from small nozzles What happens when the Glass Bulb begins to emit steam ? Reaction = Bulb Spins Action = Ejects Steam Same Principle causes a Lawn Sprinkler to Turn. THE REACTION FORCE TO THE EJECTION OF MASS CAUSES THE OBJECT TO SPIN. THIS IS THE SAME AS THE CO2 ROCKET IN THAT MATERIAL IS EXPELLED AT HIGH VELOCITY Week 3, Physics 214 Spring 2010 35 1J-04 Scale Paradox 1 A Scale Measures the Force acting on it What is the reading on the scale ? NOW, What is the reading on the scale ? WALL mg mg mg T T T T mg T = mg mg T = mg mg mg THE TENSION IN THE CORD IS THE SAME FOR BOTH CASES. THE SCALE MEASURES THE TENSION IN THE CORD. FOR EXAMPLE THE TENSION IN A ROPE IS THE SAME IF TWO PEOPLE PULL ON EACH END WITH FORCE F OR IF ONE PERSON PULLS WITH FORCE F TO A ROPE TIED TO A WALL. Week 3, Physics 214 Spring 2010 mg 36 Questions Chapter 4 Q8 A 3-kg block is observed to accelerate at a rate twice that of a 6-kg block. Is the net force acting on the 3-kg block therefore twice as large as that acting on the 6-kg block? Explain. The net force is the same Q9 Two equal-magnitude horizontal forces act on a box as shown in the diagram. Is the object accelerated horizontally? Explain. -F F No the net force is zero Q10 Is it possible that the object pictured in question 9 is moving, given the fact that the two forces acting on it are equal in size but opposite in direction? Explain. Yes, constant velocity Week 3, Physics 214 Spring 2010 37 Ch 4 CP6 A 60kg person accelerating down at 1.4m/s2 a) What is the true weight? b) What is the net force? c) What is N? d) What is the apparent weight? e) a, b, c, d with 1.4m/s2 up? a) True weight = mg N 1.4 m/s2 Mg = 60 x 9.8 = 588 N b) Net Force = Ma = 84 N c) N = 588 – 84 = 504 N d) 504 N e) ↑1.4 m/s2 Net Force = 84 N↑ Week 3, N = 588 + 84 = 672 N Physics 214 Spring 2010 W = 672 N 38 Q18 The acceleration due to gravity on the moon is approximately one-sixth the gravitational acceleration near the earth’s surface. If a rock is transported from the earth to the moon, will either its mass or its weight change in the process? Explain. It’s mass will not change but it’s weight wil be 6 times less Q22 The engine of a car is part of the car and cannot push directly on the car in order to accelerate it. What external force acting on the car is responsible for the acceleration of the car on a level road surface? Explain. Fair Ftire Freaction It’s the reaction force between the tires and the road Q23 It is difficult to stop a car on icy road surface. It is also difficult to accelerate a car on this same icy road? Explain. Because of a lack of friction the wheels will skid or spin Week 3, Physics 214 Spring 2010 39 Q25 When a magician performs the tablecloth trick, the objects on the table do not move very far. Is there a horizontal force acting on these objects while the tablecloth is being pulled off the table? Why do the objects not move very far? Explain. Yes but the force acts for a very short time and the objects start to move, then when the cloth is gone friction stops them. Q30 Two masses, m1 and m2, connected by a string, are placed upon a fixed frictionless pulley as shown in the diagram. If m2 is larger than m1, will the two masses accelerate? Explain. Yes m2 will fall and m1 will rise Week 3, Physics 214 Spring 2010 • m m 2 1 40 Q31 Two blocks with the same mass are connected by a string and are pulled across a frictionless surface by a constant force, F, exerted by a string (see diagram). A. Will the two blocks move with constant velocity? Explain. B. Will the tension in the connecting string be greater than, less than, or equal to the force F? Explain. F A. They will accelerate F = ma B. The tension will be less Week 3, Physics 214 Spring 2010 41 Q33 If you get into an elevator on the top floor of a large building and the elevator begins to accelerate downward, will the normal force pushing up on your feet be greater than, equal to, or less than the force of gravity pulling downward on you? Explain. N – mg = ma but a is negative so N is smaller than mg The only force pulling you down is gravity so if you are accelerating down the force due to gravity must be larger than the reaction force N ( N is apparent weight) Week 3, Physics 214 Spring 2010 + N g mg a 42 Ch 4 E4 A 2.5kg block is pulled with a force of 80N and friction is 5N a) What is the acceleration? Net force = 75 N Week 3, 5N 2.5 kg 80 N a = 75/2.5 = 30 m/s2 Physics 214 Spring 2010 43 Ch 4 E18 A 60kg person is in an elevator With an upward acceleration of 1.2m/s2 a) What is the net force on her? b) What is the gravitational force? c) What is the normal force? a) Net Force m = 60 KG a = 1.2 m/s2 mg F = Ma = 60 x 1.2 = 72 N b) mg = 60 x 9.8 = 588 N c) N = 588 + 72 = 660 N Week 3, Physics 214 Spring 2010 44 Ch 4 CP4 A 60kg crate is lowered from a height of 1.4m and the tension is 500N a) Will the crate accelerate? b) What is the acceleration? c) How long to reach the floor? d) How fast does the crate hit the floor? a) Net Force = 60 x 9.8 – 500 = 588 – 500 g 60 kg = 88 N b) Will accelerate down a = 88/60 = 1.47 m/s2 c) d = 1/2 at2 t = 1.38s d) v = v0 + at v = 2.03 m/s Week 3, 500 N Physics 214 Spring 2010 45 Ch 4 CP4 An 80kg crate is accelerated upward by a motor pulling on the rope. The breaking tension of the rope is 1,500N. What is the MAXIMUM possible upward acceleration of the crate? (g = 9.8 m/s2) A. 4.32 m/s2 B. 6.61 m/s2 C. 8.95 m/s2 D. 11.33 m/s2 E. 15.0 m/s2 Upward Fnet = 1500N - mg = 1500N – 784N = 716 N F = ma Week 3, so 1,500 N max. g 80 kg a = Fnet/m = 716 N / 80 kg = 8.95 m/s2 Physics 214 Spring 2010 46 Ch 4 E14 A 4kg rock is dropped and experiences air resistance of 15N a) What is the downward acceleration? 4 kg 15 N Mg F = 4 x 9.8 – 15 F = ma = 24.2 N a = 24.2/4 = 6.05m/s2 Week 3, Physics 214 Spring 2010 47 Review Chapters 1 - 4 - d + x Units----Length, mass, time SI units m, kg, second Coordinate systems Average speed = distance/time = d/t Instantaneous speed = d/Δt Vector quantities---magnitude and direction Magnitude is always positive Velocity----magnitude is speed Acceleration = change in velocity/time =Δv/Δt Force = ma Newtons Week 3, Physics 214 Spring 2010 48 Conversions, prefixes and scientific notation giga 1,000,000,000 109 billion 1 in 2.54cm mega 1,000,000 106 million 1cm 0.394in kilo 1,000 103 thousand 1ft 30.5cm centi 1/100 0.01 10-2 hundredth 1m 39.4in milli 1/1000 0.00 1 10-3 thousandth 1km 0.621mi 1mi 5280ft 1.609km 1lb 0.4536kg g =9.8 1kg 2.205lbs g=9.8 micro 1/1,000,000 1/106 10-6 millionth nano 1/1,000,000,000 1/109 10-9 billionth Week 3, Physics 214 Spring 2010 3.281ft 49 Speed, velocity and acceleration v = Δd/Δt a = Δv/Δt The magnitude of a is not related to the magnitude of v the direction of a is not related to the direction of v 2 3 4 1 v = v0 + at constant acceleration d = v0t + 1/2at2 d,v0 v,a can be + or d = 1/2(v + v0) t v2 = v02 + 2ad Week 3, Physics 214 Spring 2010 50 One dimensional motion and gravity v = v0 + at d = v0t + 1/2at2 v2 = v02 + 2ad d = ½(v + v0)t + g = -9.8m/s2 + Week 3, At the top v = 0 and t = v0/9.8 At the bottom t = 2v0/9.8 Physics 214 Spring 2010 51 Equations v = v0 + at d = v0t + 1/2at2 d = ½(v + v0)t v2 = v02 + 2ad Sometimes you have to use two equations. NO YOU DON’T! 4th eqn. has already eliminated t, answer in one step. v0 = 15m/s v = 50m/s What is h? v = v0 + at v0 50 = 15 + 9.8t t = 3.57 s ` h = v0t + 1/2at2 g h 2 v Week 3, h = 15 x 3.57 + 1/2x9.8x3.57 = 116m h = ½(15 + 50) x 3.57 = 116m Physics 214 Spring 2010 52 Projectile Motion axis 1 axis 2 v1 = constant and d1 = v1t vv = v0v + at and d = v0vt + 1/2at2 v1 g 9.8m/s2 h v R Use + down so g is + and h is + h = v0vt + 1/2at2 v0v = 0, t2 = 2h/a R = v1t v = v0v + at Week 3, Physics 214 Spring 2010 53 Complete Projectile v0v v1 9.8m/s2 v1 v1 v0v highest point the vertical velocity is zero vv = v0v + at so t = v0v/9.8 h = v0vt + 1/2at2 end t = 2v0v/9.8 and R = v1 x 2v0v/9.8 and the vertical velocity is minus v0v Week 3, Physics 214 Spring 2010 54 Newton’s Second and First Law Second Law F = ma unit is a Newton (or pound) First Law F = 0 a = 0 so v = constant Third law For every force there is an equal and opposite reaction force N Weight = mg mg Ff F F Ff F = ma Week 3, v = v0 + at d = v0t + ½ at2 d = ½(v + v0)t v2 = v02 + 2ad Physics 214 Spring 2010 55 Examples + T N g 30 – 8 – T = 4a T – 6 = 2a 30 – 8 – 6 = 6a mg N – mg = ma a + N > mg a – N < mg Week 3, Physics 214 Spring 2010 56