MOTION, FORCES & ENERGY TAKS REVIEW IPC (4)

IPC (4) The student knows the concepts of force and motion evidence in everyday life. (A) The student is expected to calculate speed, momentum, acceleration, work and power in systems such as in the human body, moving toys and machines.

SPEED / VELOCITY To calculate speed/velocity (s), we need 1. Distance traveled (d) in meters (m) & 2. Time (t) in seconds (s) Our formula is: Speed = distance traveleds = d time t The units for speed are usually m/s. THIS IS ON YOUR FORMULA SHEET!

Problem 1: speed / velocity A car traveled 150 km in 2.5 hours. What was its average speed in km per hour? Identify the information given (look at units): d=150 km, t=2.5 hrs Select the formula: s=d/t Plug in the numbers and solve: s = 150km / 2.5 hrs Use the calculator to obtain an answer with correct units: s = 60 km/hr

Problem 2: speed / velocity The diagram represents the total travel of a teacher on a Saturday. Which part of the trip is made at the greatest average speed? This problem requires four steps. We need to calculate the speed for each part of the trip. s (Q) = 14/12 = 1.2 km/min s (R) = 12/8 = 1.5 km/min s (S) = 15/9 = 1.7 km/min s (T) = 11/15 = 0.7 km/min The part with the greatest average speed is: S (1.7 km/min).

Problem 3: speed GRAPHS The slope of a distance-time graph is the speed. The graph shows the distance traveled by a vehicle over a certain period of time. Which segment of the graph shows the vehicle moving with the greatest speed? On a speed graph, the part of a line with the steepest slope has the greatest speed. ANS: L Which segment represents the vehicle at rest? “At rest” means the speed is zero. A line with a slope of 0 is horizontal.

MOMENTUM Momentum measures how much force is necessary to stop an object from moving. The heavier and/or faster an object is, the more momentum it has. large mass large velocity large momentum small mass small velocity small momentum

MOMENTUM To calculate momentum (p), we need 1. Mass (m) in kilograms (kg) & 2. Velocity (v) in meters per second (m/s) Our formula is: Momentum = mass x velocity p = mv So the units for momentum are kg●m/s. THIS IS ON YOUR FORMULA SHEET!

Problem 4: momentum The pictures show how an air bag functions in a collision. How much momentum in kg●m/s does the airbag absorb from the crash-test dummy if all the crash-test dummy’s momentum is absorbed by the air bag? p = mv = 100 kg (6.3 m/s) = 630 kg●m/s Car velocity = 6.3 m/s Driver velocity = 6.3 m/s Driver mass = 100 kg Car velocity = 0 m/s Driver velocity = 6.3 m/s Driver mass = 100 kg Car velocity = 0 m/s Driver velocity = 0 m/s Driver mass = 100 kg

Problem 5: momentum Which bike rider has the greatest momentum? A. A 40 kg person riding at 45 km/h p = mv = 40 kg (45 km/h) = 1800 kg●m/s B. A 50 kg person riding at 35 km/h p = 50 kg (35 km/h) = 1750 kg●m/s C. A 60 kg person riding at 25 km/h p = 60 kg (25 km/h) = 1500 kg●m/s D. A 70 kg person riding at 15 km/h p = 70 kg (15 km/h) = 1050 kg●m/s

ACCELERATION To calculate acceleration (a), we need 1. Initial and final velocities ( V F & V I ) in meters per second (m/s) & 2. Change in time ( ∆t) in seconds (s) Our formula is: Acceleration = Final v – Initial v / change in time a = v F -v I / ∆t So the units for acceleration are m/s 2. THIS IS ON YOUR FORMULA SHEET!

Problem 6: acceleration According to this graph, what was the bicycle’s acceleration between 10 and 12.5 seconds? At 10 s, v I is 6.5 m/s. At 12.5 s, v F is 8.5 m/s. Δt = 12.5 s -10 s = 2.5 s a = v F – v I / Δt = 8.5 m/s – 6.5 m/s / 2.5 s a= 0.8 m/s 2

WORK To calculate work (W), we need 1. Force (f) in Newtons (N) & 2. Distance (d) in meters (m) Our formula is: Work = force x distance W = fd So the units for work are N●m or Joules (J). THIS IS ON YOUR FORMULA SHEET!

POWER To calculate power (P), we need 1. Work (W) in Joules (J) & 2. Time (t) in seconds (s) Our formula is: Power = work / time P = w / t So the units for power are J/s or Watts (W). THIS IS ON YOUR FORMULA SHEET!

Problem 7: work & power The weight lifter used a force of 980 N to raise the barbell over her head in 5.21 seconds. Approximately how much work did she do in raising the barbell? W = Fd = 980 N (5.21 s) = 5106 J How much power did she have? P = W / t = 5106 J / 5.21 s P = 980 W

Problem 8: work & power If a force of 100 newtons was exerted on an object and no work was done, the object must have — A. accelerated rapidly B. remained motionless C. decreased its velocity D. gained momentum

Problem 9: work & power A horizontal force of 600 N is used to push a box 8 m across a room. Which of these variables must be known to determine the power used in moving the box? A. The weight of the box B. The potential energy of the box C. The time it takes to move the box D. The length of the box

FORCES The overall force on an object is called the net force. An object will accelerate when there is a net force acting on it. Friction is the name given to the force that acts between materials that are in contact and moving past each other. Air resistance is friction between objects and the air. Gravity is the force that gives an object its weight, pulling it toward the center of the earth at an acceleration of 9.8 m/s 2.

Newton’s 1 ST Law An object at rest will stay at rest and an object in motion will continue that motion… (this is called INERTIA) …unless an unbalanced force acts on it. If you are in an accident and aren’t wearing a safety belt you would continue your motion… Do not let this happen to you; buckle up!

Newton’s 2 ND Law The unbalanced force acting on an object equals its mass times its acceleration. Force = mass x acceleration Ex: A filled wheelbarrow requires more force to move than an empty one. The acceleration of a golf ball times its mass is equal to the force with which it’s hit.

Action / Reaction For every action there is an equal and opposite reaction. What is the action? What is the reaction? The gases push downward out of the rocket, The rocket is pushed upward by gases. Newton’s 3 RD Law

Problem 10: Forces A catapult was designed to project a small metal ball at a target. The resulting data are shown in the table. Which of these might explain the difference between the calculated and actual distances? A The ball landed short of the calculated distance because of an increase in momentum. B Air resistance caused the ball to land short of the calculated distance. C Initial mass of the ball changed with each trial. D The metal ball was too small for accurate measurements to be made.

Problem 11: Forces After shooting a cannonball, a cannon recoils with a much lower velocity than the cannonball. This is primarily because, compared to the cannonball, the cannon has a — A. much greater mass B. smaller amount of momentum C. greater kinetic energy D. smaller force applied to it

The frog leaps from its resting position at the lake’s bank onto a lily pad. If the frog has a mass of 0.5 kg and the acceleration of the leap is 3 m/s 2, what is the force the frog exerts on the lake’s bank when leaping? According to Newton’s 2 nd Law, F = ma F = (0.5 kg)(3 m/s 2 ) F = 1.5 N Problem 12: Forces

Problem 13: Forces The illustration to the right shows a student about to throw a ball while standing on a skateboard. Which illustration below correctly shows the skateboard’s direction of motion after the student releases the ball? ABCD

Machines Pulley: a rope around a grooved wheel. Used to change the direction of the force applied to it. Ex: flagpole, block & tackle Lever: a bar that pivots around a fixed point (fulcrum). Generally, the farther the effort (input) force is away from the fulcrum, and the closer the load is to the fulcrum, the easier it is to lift an object.

Remember! Machines do not reduce the amount of work done; they only change the amount of force, the distance, the direction or the speed. The efficiency of a machine tells us the % of work put into the machine that is useful. You can never get more work out of a machine than you put into it! (Efficiency can never be > 100%) Machines

Which lever arrangement requires the least effort force to raise a 500 N resistance? Problem 14: machines A. D. C. B.B.

What is the efficiency of an air conditioner if there is a work input of 320 J and a work output of 80 J? work output (W O ) = 80 J work input (W I ) = 320 J % EFF = W O x 100% = 80 J x 100% = 25% W I 320 J Problem 15: machines