HONORS PHYSICSGood morning! Please…• Take out your homework, to be checked• Answer the warm-up question: • A baseball is thrown straight down from a building that is 65 meters tall. Its initial velocity is 3.0 m/s. What will its velocity be right before it hits the ground?
AGENDA:1. Physics in the News!2. Finish projectile motion activity3. Review4. Review activitiesNOTE: Midterm is Wednesday 9:45-11:45- Bring a book or work to do if you finish the midterm early• LAST DAY FOR HOT TASK/MIDTERM WILL BE WEDNESDAY.• **Cheat sheet update***
MIDTERM TOPICSTopics Suggested Study Techniques1. Conversion, measurement, precision, • Look over all old tests & know how to accuracy do every problem2. One-dimensional motion: constant • Go over your notes from each unit – speed be familiar with concepts3. One-dimensional motion: accelerated • Re-do problems from the homework & motion problems from the review sheet4. Free-fall motion Format:5. Vectors • 30 multiple choice6. Projectile motion • 14 problems with multiple parts7. Forces & Newton’s laws
1. CONVERSION, MEASUREMENT, PRECISION,ACCURACY• Using factor label method to convert numbers• What are some examples of SI units used in physics? English units?• What are the significant figures for the following two numbers: • 0.02003 54300• What is precision? What is accuracy?• How to convert between metric units • Know kilo- to milli-
2 & 3. ONE-DIMENSIONAL MOTION: CONSTANTSPEED & ACCELERATION• Graphing motion • Identifying the information given on an x vs. t, v vs. t, a vs. t graph • Slope of the lines on the graphs • Area under the curve• Solving problems using one-dimensional motion equations
4. FREE-FALL MOTION• The acceleration of an object in free-fall• Using motion equations to solve free-fall problems: • An object dropped from rest • An object dropped with an initial velocity • An object thrown straight up in the air that comes back down.
4. FREE-FALL MOTION• Sample problem: a ball is thrown straight up into the air at 12.0 m/s. • What is the velocity of the ball at its maximum height? What is its acceleration at maximum height? 7.4 m • What is the maximum height that the ball will reach? • What will be the velocity of the ball after 2.0 seconds? -7.6 m/s
5. VECTORS• Difference between scalar and vector quantities • What are some examples of each?• Breaking vector into its components• Solving for resultant vectors by adding or subtracting two vectors
6. PROJECTILE MOTION• Concepts behind projectile motion • Projectile motion is anything that has an initial horizontal velocity that is under the influence of free-fall • X & y dimensions of motion are completely independent of one another • What happens in the x-direction? • What happens in the y-direction?• What is range? At what angle does any object have maximum range?• Problems with initial horizontal velocity• Problems with projectiles launched at an angle.
7. FORCES• Newton’s laws and how they apply in various situations • What is Newton’s first law? Second law? Third?• Drawing free-body diagram• Problems with friction on an inclined plane• Problems with friction on a level surface• Problems with a force applied at an angle• Calculating coefficients of kinetic and static friction.
7. FORCES• A student moves a box of books down the hall by pulling on a rope attached to the box. The student pulls with a force of 185 N at an angle of 25.0 degrees above the horizontal. The box has a mass of 35.0 kg and the coefficient of friction between the box and floor is 0.27. Find the acceleration of the box. -1.2 m/s2
7. FORCES • What is the net force on each object? FN = 10 NFf = 3 N Fa = 5 N Mg = 10 N FN = 10 N FN = 10 N Ff = 3 N Fa = 1 N Ff = 3 N Fa = 3 N Mg = 10 N Mg = 10 N
HOT TASK• What does the normal force do? 4.0 kg• How can we find the normal in both situations shown? 15o 15o
NEWTON’S ACTIVITIES• When a constant force is applied to something, should the speed increase? Should it stay the same? • Use Newton’s Second law and think about the scooter activity – did speed change as force was constant? Did acceleration change?• Explain how Newton’s first law applies in a car when wearing a seatbelt vs. without a seatbelt.
Example: force applied at an angleSuppose a 10.0 kg box is pulled at an angle of 30.0 degrees with a force of 50.0 Newtons. F ¹ mg! Na) Calculate the normal force of the box å FY = 0 ® FN + Fay - Fg = 0 FN + Fay = mg FN = mg - Fay ® (10)(9.8) - 50sin30 FN = 73Na) If the block is accelerating at 1.5 m/s/s, calculate the coefficient of friction. FN Fa = 50 N å Fx = Fnet = ma Fay = 50sin(30) Fax - Ff = ma Ff 30 Fax = 50cos(30) 50 cos30 - m FN = (10)(1.5) mg 43.3- m (73) = 15 m = 0.39
EXAMPLE: FORCE APPLIED AT ANGLE#5 from handout last week:A 120.0 N force is applied at an angle of 45 degrees to a 40.0 kg box. The coefficient offriction between the two surfaces is 0.15. Find(b) The force due to gravity and normal force(c) The force due to friction(d) The acceleration(e) How fast the object is moving in 10.0 sec(f) How far it will move in 10.0 sec
NEWTON’S THIRD LAW“For every action there is an EQUAL and OPPOSITE reaction. • This law focuses on action/reaction pairs (forces) • They NEVER cancel out All you do is SWITCH the wording! •PERSON on WALL •WALL on PERSON
NEWTON’S THIRD LAW• How does this law apply when there is a collision between two objects, like a train and a truck??
NEWTON’S THIRD LAW This figure shows the force during a collision between a truck and a train. You can clearly see the forces are EQUAL and OPPOSITE. To help you understand the law better, look at this situation from the point of view of Newton’s Second Law. Ftruck = Ftrain What about mass & acceleration just after collision? mtruck Atruck = Mtraina trainThere is a balance between the mass and acceleration. One object usuallyhas a LARGE MASS and a SMALL ACCELERATION, while the other has aSMALL MASS (comparatively) and a LARGE ACCELERATION.
N.T.L EXAMPLES Action: HAMMER HITS NAIL Reaction: NAIL HITS HAMMER Action: Earth pulls on YOU Reaction: YOU pull on the earth
NEWTON’S FIRST LAWConcepts Problems• What is it? • What type of problems apply? • An object in motion will remain in motion, or an object at rest will remain at rest unless acted upon 4.0 kg by an unbalanced force.• Fnet = 0 • What conditions can occur? 4.0 kg • Object can be at rest • Object can be in constant motion at a constant speed
NEWTON’S SECOND LAWConcepts Problems• What is it? • What type of problems apply? • Fnet = ma or re-arranged as.. a = Fnet 4.0 kg m• Fnet = ma • Describe the motion of the object when Newton’s 2 nd law applies 4.0 kg
ATWOOD’S MACHINE Draw the following FBDs if mass 1 = 1.0 kg, mass 2 = 2.0 kg: • An FBD of the entire system • An FBD of mass 1 • An FBD of mass 2 • Which direction will the masses move? • Use the FBDs to find the net force equations for each FBD • Calculate the acceleration
FREE-BODY DIAGRAMS• Draw a Free-body diagram of an elevator accelerating up, and then one of an elevator accelerating down. Write the equation for the net force on the elevator for each.
FREE-BODY DIAGRAM• A block is pushed across a frictionless plane, and released when the block reaches a speed of 2.0 m/s. Draw the FBD of the block AFTER it is released. 4.0 kg
HOMEWORK24. A 650 N force acts in a northwesterly direction. A second 650-N force must be exerted inwhat direction so that the resultant of the two forces points westward?
HOMEWORK*26. Sketch the free-body diagram of a baseball (a) at the moment it is hit by the bat, (b) afterit has left the bat and is flying toward the outfield
HOMEWORK27. USE figure from book: two forces F1 and F2 act on a 27.0 kg object on a frictionlesstabletop. If F1 = 10.2 N and F2 = 16.0N, what is the net force on the object and itsacceleration in both situation (a) and (b) F1 F2
HOMEWORK32. A window washer pulls herself upward using the bucket-pulley apparatus shown. Mass =65 kg. (a) how hard must she pull downward to raise herself slowly at constant speed? (b) If she increases this force by 10%, what will her acceleration be?
LAST NIGHT’S HOMEWORK23. a) 40 N, b) 10 N, c) 038. 100 N, no force*39. (a) μs=0.82 (b) μk=0.7440. diagrams41. μs=0.4142. μs=2.3
HOMEWORK: FRICTION PROBLEMS38. The coefficient of Friction between a 35 kg crate and the floor is 0.30, what horizontalforce is required to move the crate at a steady speed across the floor? What horizontal forceis required if μk is zero? (100 N, no force)
HOMEWORK: FRICTION*39. A force of 40.0 N is required to start a 5.0 kg box moving across a horizontal concretefloor. (a) what is the coefficient of static friction between the box and the floor? (b) if the 40.0N force continues, the box accelerates at 0.70 m/s2. What is the coefficient of kinetic friction?(a) μs=0.82 (b) μk=0.74 FNFf Fa mg
HOMEWORK: FRICTION*40. (a) a box sits at rest on a rough 30 degree inclined plane. Draw the free-body diagram,showing all the forces cting on the box.(b) How would the diagram change if the box were sliding down the plane?(c) How would it change if the box were sliding up the plane after an initial shove?
HOMEWORK: FRICTION41. A 2.0-kg silverware drawer does not slide readily. The owner gradually pulls with more andmore force. When the applied force reaches 8.0 N, the drawer suddenly opens, throwing allthe utensils to the floor. Find the coefficient of static friction between the drawer and thecabinet. (μs=0.41)
HOMEWORK: FRICTION42. Drag race tires in contact with an asphalt surface probably have one of the highestcoefficients of static friction in the everyday world. Assuming a constant acceleration and noslipping of tires, estimate the coefficient of static friction for a drag racer that covers thequarter mile in 6.0 s. (μs=2.3)
CALCULATING COEFFICIENTS OF STATIC & KINETIC FRICTIONYou can calculate μs and μk between two surfaces Materials needed:by… • Brick of known mass with string• Finding out the MAXIMUM force required to attached begin moving an object • Spring scale Fmax = μsFN• Finding out the force required to move an object at a constant speed Directions: • Have TWO different people take readings F f = μkFN from the force scales & take the average of the two In both of these cases, the object is in equilibrium, so • Use equations to solve for coefficients • NOTE: Spring scale reads in Pounds – Fa = Ff & Fa = Fmax you will have to conver units to Newtons (using 2.2 lbs = 1 kg, W = mg)
FORCES ON AN INCLINED PLANEFf = μFN FNSo to find Ff, we must find FN first!Steps to solving these problems…1. FBD2. Sum forces in y-direction (to find Fn) FG3. Solve for Ff4. Find the net force in the x- θ direction (direction of motion)
WORKSHEET PROBLEM #1 12 kg1. A 12-kilogram block is place on a 25 o inclined plane. 25o s= 0.15 and k= 0.11(a) Draw a Free Body Diagram showing mg, F , FN , FII & Ff.(b) Calculate for mg, FII , FN for the block.(c) Calculate the forces of static & kinetic friction.(d) Solve for the net force on the block.(e) Solve for the acceleration of the block.(f) How far will it slide in 1.0 seconds?
WORKSHEET PROBLEM #1 12 kg1. A 12-kilogram block is pushed up a 25o inclinedplane, then the applied force is removed. Answer the 25ofollowing questions for the block AFTER it is released,but while it is still traveling UP the incline. s= 0.15 and k= 0.11(a) Draw a Free Body Diagram showing mg, F , FN , FII & Ff.(b) Calculate for mg, FII , FN for the block.(c) Calculate the forces of static & kinetic friction.(d) Solve for the net force on the block.(e) Solve for the acceleration of the block.(f) How far will it slide up the incline if the initial velocity is 2.0 m/s2?
INTRO TO FRICTION!What do you already know about friction?When is friction useful?When is it harmful?Write a short paragraph about a friction-free world!
FRICTIONAL FORCES • There are two types of frictional forces: Type of friction Definition EquationStatic Friction A force that acts parallel to the two surfaces & Fmax = μsFN(used for stationary objects) keeps an object from moving.Kinetic Friction A force that acts opposite to the direction of an(used for moving objects) Ff = μkFN object’s motion. Fmax is the maximum force that can be applied to an object before it Ff is the frictional force begins to move The greek symbol “μ” is pronounced “mu”
NEW FREE-BODY DIAGRAM• Now we include the force of friction! FN Ff Fa mg
HOW TO CALCULATE COEFFICIENTS OFFRICTION (WHEN NET FORCE = 0) • When an object is moving at a CONSTANT SPEED, we can find the force of friction due to the coefficient of kinetic friction. Constant speed means that acceleration = 0 FN So Fnet = 0 SO…. Fa – Ff = 0Ff Fa Ff = Fa mg • When a force is applied that causes an object to JUST BEGIN TO MOVE, we can find the force of friction due to the coefficient of static friction Once again…. Ff = Fa
Example: Net force = 0A 10-kg box is being pulled across the table to the right at a constant speed with a force of 50N.a) Calculate the Force of Friction Fa = Ff = 50Na) Calculate the Force Normal mg Fn (10)(9.8) 98Na) Calculate the coefficient of kinetic friction Ff = mk FN FN Fa 50 = m k (98) Ff 50 mg m k = = 0.51 98
Example: Net force = 0 Suppose the same box is now pulled at an angle of 30 degrees above the horizontal. a) Calculate the Force of Friction Fax Fa cos 50cos30 43.3N Ff Fax 43.3N a) Calculate the Force Normal FN m g! FN Fay mg FN Fa FN m g Fay (10)(9.8) 50 sin 30 Fay FN 73NFf 30 Fax mg
EXAMPLE: NET FORCE = MAA 50 N applied force drags an 8.16 kg log to the right across a horizontal surface. What is the acceleration of the log if the force of friction is 40.0 N? Fn a FNET = ma 50 N40 N Fa - Ff = ma mg 50 - 40 = 8.16a 10 = 8.16a a= 1.23 m/s/s
EXAMPLE: NET FORCE = MAA sled is being accelerated to the right at a rate of 1.5 m/s/s by a rope at a 33 degree angle above the + x . Calculate the acceleration of the sled if the Frictional Force is 26.8 N, the mass of the sled is 66 kg and the tension in the rope is 150 N. a FN Tsin Tcos FNET = ma Ff T cosq - Ff = ma mg T cosq - ma = Ff 150 cos33- (66)(1.5) = Ff Ff = 1.5 m/s/s
An Atwoods machine is a device where two masses, m2 and m1, are connected byATWOOD’S MACHINE a string passing over a pulley. • Assume the pulley is frictionless and massless, which means the tension is the same everywhere in the string. • To solve these problems: • Make three FBDs: one for each mass, and one for the overall system. • So you can make 3 Fnet=ma equations. • To find Tension: use the FBDs of individual masses • To find acceleration: use either both FBDs of individual masses, or one of the overall system
If mass 1 is 200 kg, and mass 2 is 30 kg, what is the acceleration of the system?ATWOOD’S MACHINE a = 3.27 m/s2• Make FBDs• Have the direction of acceleration be positive• Use Newton’s 2 nd law to find acceleration
HOMEWORK SOLUTIONS Pg. 104 # 1-81. 69 N Free Body Diagrams2. 116 kg3. 883 N a. A projectile in motion in the presence of air resistance.4. 1260 N5. (a) 648 N b. A car at the instant it hits a brick wall. (b) 112 N c. A heavy crate being pushed across a (c) 244 N surface (neglect surface friction). (d) 0 N d. A shopping cart being pushed at a 30°6. (a) W = 196 N, Fn = 196 N angle with horizontal (neglect surface friction). (b) Fn on 20 kg box: 294 N, Fn on 10 kg from20 kg = 98 kg7. 3443 N8. F = 153 N