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Motion. Chapter 11. Standards. Students will: SPS8. Determine the relationship between force, mass and motion SPS8a Calculate velocity and acceleration SPS8c Relate falling objects to gravitational force SPS8d Explain difference in mass and weight. Observing Motion.
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Motion Chapter 11
Standards • Students will: • SPS8. Determine the relationship between force, mass and motion • SPS8a Calculate velocity and acceleration • SPS8c Relate falling objects to gravitational force • SPS8d Explain difference in mass and weight
Observing Motion • Motion-change in position in relation to a reference point.
Measuring Motion: Distance • Distance- how far an object moves on a path • Displacement- how far between starting and ending points on a path
Measuring Motion: Speed • Speed • rate of motion • distance traveled per unit time speed = distance time
Measuring Motion: Speed (cont’d) • Instantaneous Speed • speed at a given instant • Average Speed- = total distance total time
Measuring Motion: Velocity • Velocity • speed in a given direction • can change even when the speed is constant!
Graphing Speed/Velocity • X axis- usually independent variable (time) • Y axis- usually dependent variable (distance) • Slope of straight line= vertical change horizontal change
Graphing (cont’d) • slope = velocity • steeper slope = faster velocity • straight line = constant velocity • flat line = 0 velocity (no motion)
Calculating Slope Distance Vs Time 16 • Choose two points on graph to calculate slope. • Calculate the vertical and horizontal change. • Divide the vertical change by the horizontal change. 14 . 12 10 Distance (m) 8 . 6 4 2 0 1 2 3 4 5 Time (s)
Calculating Slope(cont’d) Distance Vs Time 16 1. point 1: d= 6m t= 1s point 2: d= 12m t= 4s 2. vert ∆- 12m-6m= 6m horiz ∆- 4s-1s= 3s 3. slope= 6m= 2m/s 3s 14 . 12 10 Vertical change Distance (m) 8 . 6 Horizontal Change 4 2 0 1 2 3 4 5 Time (s)
Practice Problem • Who started out faster? • A (steeper slope) • Who had a constant speed? • A • Describe B from 10-20 min. • B stopped moving • Find their average speeds. • A = (2400m) ÷ (30min) A = 80 m/min • B = (1200m) ÷ (30min) B = 40 m/min A B
Measuring Motion: Acceleration • Acceleration • the rate of change of velocity • change in speed or direction • Centripetal acceleration • circular motion (even if speed is constant, direction is always changing) ex.Moon accelerates around Earth
Measuring Motion: Acceleration • Positive acceleration -“speeding up” • Negative acceleration -“slowing down”
Calculating Acceleration • a= vf – vi t • a= ∆ v t • a- acceleration • vf- final velocity • vi- initial velocity • t- time
Calculating Acceleration (cont’d) • List given, then unknown values. • Write equation for acceleration. • Insert known values into equation and solve. Vf-Vi a t t
Practice Problem 1 A flowerpot falls off a second-story windowsill. The flowerpot starts from rest and hits Mr. Mertz 1.5s later with a velocity of 14.7m/s. Find the average acceleration of the flowerpot. Given:Remember:Solve: t= 1.5s a= vf – vi a= 14.7m/s-0m/s Vi= 0m/s t 1.5s Vf= 14.7m/s a= 14.7m/s a= ? 1.5s a= 9.8m/s2
Practice Problem 2 Joseph’s car accelerates at an average rate of 2.6m/s2. How long will it take his car to speed up from 24.6m/s to 26.8m/s2? Given:Remember: Solve: a= 2.6m/s2 t= (vf-vi) ÷ a vf= 26.8m/s2 vf-vi t= 26.8m/s2-24.6m/s2 Vi= 24.6m/s2 a t 2.6m/s2 t= ? t= 2.2m/s 2.6m/s2 t= 0.85s
Practice Problem 3 A cyclist travels at a constant velocity of 4.5m/s westward and then speeds up with a steady acceleration of 2.3m/s2. Calculate the cyclist’s speed after accelerating for 5.0s. Given: Remember: Solve: vi= 4.5m/s vf= vi + a x t vf= ? vf-vi vf= 4.5m/s + (2.3m/s2 x 5.0s) a= 2.3m/s2 a t vf= 4.5m/s + 11.5m/s t= 5.0s vf= 16m/s
Graphing Acceleration Distance/Time On Distance-Time graph: • Acceleration is shown as a curved line
Graphing Acceleration Speed/Time On a Speed-Time graph: • Slope of straight line= acceleration • Positive slope- speeding up • Negative slope- slowing down • Flat line- constant velocity (no acceleration)
Newton’s Laws of Motion • 2nd Law of Motion: The acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. Force = mass x acceleration or or F= ma
Force Force- push or pull that one body exerts on another
Fundamental Forces • 4types: 1. gravity 2. electromagnetic 3. weak nuclear 4.strong nuclear • Vary in strength • Act through contact or at a distance
Forces • Force Pairs: forces moving in opposite directions • Balanced forces: do not move; push equally on each other • Unbalanced forces: acceleration (movement) in the direction of larger force
Friction Friction: forcethat opposes motion between 2 surfaces • Static Friction: nonmoving surfaces • Kinetic Friction: moving surfaces- sliding or rolling (sliding friction is greater than rolling friction)
Friction Facts • Necessary for all motion • Rougher surfaces create greater friction • Greater mass creates greater friction • Lubricants reduce friction
Newton’s First and Second Laws Chapter 12
Newton’s Laws of Motion • 1st Law of Motion: Law of Inertia • An object at rest will remain at rest; • An object in motion will continue moving at a constant velocity unlessacted upon by a net force.
Inertia • Objects move only when net force is applied. • Objects maintain state of motion. • Inertia is related to mass. (small mass can be accelerated by small force large mass can be accelerated by large force)
Newton’s Laws of Motion • 2nd Law of Motion: The acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. Force = mass x acceleration or or F= ma
Newton’s Second Law • For equal forces, large masses accelerate less • Force is measured in newtons (N) • 1N= 1kg x 1m/s2 F m a
Practice Problem 1 Zoo keepers lift a stretcher that holds a sedate lion. The lion’s mass is 175 kg and the upward acceleration of the lion and stretcher is 0.657m/s2. What force is needed to produce acceleration of the lion and stretcher? Given: Remember: Solve: m= 175 kg F = m x a a= 0.657m/s2 F = 175 kg x 0.657m/s2 F= ? m a F = 11.49 N F F
Practice Problem 2 A baseball accelerates downward at 9.8 m/s2. If gravity is the only force acting on the baseball and is 1.4N, what is the baseball’s mass? Given: Remember: Solve: F= 1.4 kg/m/s2 m = F a= 9.8 m/s2 a m= ? m = 1.4 N 9.8 m/s2 m = .14 kg F m a
Weight and Mass • Weight- measure of gravity on an object • Not equal to mass (constant everywhere) • Measured in Newtons weight = mass xfree-fall acceleration (9.8m/s2) free-fall m g acceleration w
Gravity • Force of attraction between 2 objects in the universe • Increases as: - mass increases - distance decreases • Affects all matter
Gravity Quiz 1 Who experiences more gravity - the astronaut or the politician? More distance Less distance
Gravity Quiz 2 Which exerts more gravity, your hand or your pencil ?
Gravity Quiz 3 • Would you weigh more on Earth or Jupiter? (Hint: Which planet has the greater mass?)
Free Fall Acceleration • Occurs when Earth’s gravity is only force acting on an object • In absence of airresistance, all objects accelerate at same rate • g = 9.8 m/s2 • g = 9.8 m/s2 g = 9.8 m/s2 g = 9.8 m/s2 g = 9.8 m/s2 g = 9.8 m/s2
Air Resistance • Force of air on a moving object which opposes its motion • Aka fluid friction or drag • Depends on objects: - speed - surface area - shape - density
Air Resistance (cont’d) • Terminal velocity= maximumvelocity reached by a falling object • Reached when… F gravity = F air resistance (no net force)
Projectile Motion Projectile- • Any object thrown in air • Only acted on by gravity • Follows parabolic path- trajectory • Has horizontal and vertical velocities V oy V ox
Projectile Motion (cont’d) • Horizontal Velocity • depends on inertia • remains constant • Vertical Velocity • depends on gravity • accelerates downward at 9.8 m/s2
Projectile Motion Quiz • A moving truck launches a ball vertically (relative to the truck). If the truck maintains a constant horizontal velocity after the launch, where will the ball land (ignore air resistance)? A) In front of the truck B) Behind the truck C) In the truck Answer: C because horizontal and vertical velocities are independent of each other
Newton’s Laws of Motion • 3rd Law of Motion: When one object exerts a force on a second object, the second object exerts an equal but opposite force on the first.
Newton’s Third Law • Forces always occur in pairs • Forces in a pair do not act on the same object • Equal forces don’t always have equal effects
Newton’s Third Law Problem: How can a horse pull a cart if the cart is pulling with equal force back on the horse?