UNIT4: FORCES, MOTIONS AND ENERGY MODULE 1 : FORCES AND MOTION FRANCISCA L. DITAN TEACHER SCIENCE 9
WHAT IS A FORCE? FORCE ….A PUSH OR A PULL Mathematically it is represented by an equation: F = ma where: F = force m = mass a = acceleration Unit used: Force = kg.m /s 2 mass = kg acceleration = m/s 2
example A boy wants to push a 5kg object at a rate of 5m/s 2 . How much force was exerted by the boy to push the object? Given: 5kg ….. m 5m/s 2 …. A Required: force (F) Solution: F = ma = 5kg x 5 m/s 2 = 25 kg.m /s 2
What is a motion? Motion is the ability of the objective to move. Speed and Velocity measures the rate of motion. PHYSICAL QUANTITIES OF MEASUREMENT 1. SCALAR QUANTITY 2. VECTOR QUANTITY SCALAR QUANTITY involves magnitude only Ex. Speed . Mass , time
Vector quantity Involves magnitude and direction Ex: Velocity , acceleration . Force etc.
Uniform accelerated motion:horizontal dimension described the object to move in a straight path Examples: A vehicle in a highway moving A ball rolling
Usefull equations used in solving problems involving uniform accelerateD motion (please refer to your text book on page 235 to 236 EQUATION A: V = d/t Where : v = velocity d = distance travelled t = time Units used in velocity m/s , km/h, m/min etc.
EQUATION B: AVERAGE VELOCITY V ave = vf + vi / 2 Where : V ave = average velocity Vf = final velocity Vi = initial velocity
EQUATION C : ( ACCELERATION) a = vf – vi / t Where : a = acceleration vf = final velocity vi = initial velocity UNITS USED : m/s 2
Equation d : ( distance travelled or the displacement d = ( vf + vi) t 2 Where : d = distance or the displacement vf = final velocity vi = initial velocity t = time UNIT USED: meter(m), kilometer (km), feet(ft,) miles (mi)
EQUATION E : DISPLACEMENT OR DISTANCE d= vit + at 2 2 shows the displacement of the body is directly proportional to the square of time. It confirms that for equal interval of time, displacement increases quadraticaly .
EQUATION F: FINAL VELOCITY DEPENDS ON DISPLACEMENT Vf 2 = vi 2 + 2ad Where: Vf = final velocity Vi = initial velocity a= acceleration d= distance or displscement
Assessment: Identify the following units 1. mg 2 m 3. s 4. m/s 5. m/s 2 6. hrs 7. kg 8. km/h 9. kg.m /s 2 10.miles
summary 1. V = d/t 2. V ave = vf + vi / 2 3. a = vf – vi / t 4. d = ( vf + vi) t 2 5. d= vit + at 2 2 6. Vf 2 = vi 2 + 2ad
SAMPLE PROBLEM An airplane from rest accelerates on runway at5.50m/s 2 for 20.25 s until it finally takes off the ground. What is the distance covered before take off? Given :: 5.50 m//s2….. a 20.25 s ……… t 0m/s ………… vi Required : d Solution: d= vit + ½ at2 = ( om /s)(20.25s) + ½ ( 5.50m/s2)(20.25s)2 d = 1128m
Sample problem 2: 1. Problem 1: From rest , a car accelerated at 8 m/s 2 for 10 seconds. a) What is the position of the car at the end of the 10 seconds? b) What is the velocity of the car at the end of the 10 seconds? Given : 0m/s …… vi 8m/s 2 ……. a 10s ……… t
SOLUTION PROBLEM 2 Required: A. distance (d) B. final velocity ( Vf ) Solution : A. d= vit + ½ at 2 = 0 + ½ (8m/s 2 )(10s) 2 = 400m B . vf = vi + at = o + (8.0m/s 2 )(10s) vf = 80m/s
Sample problem 3 A jeepney from rest accelerates uniformly over a period of time of 3.25 seconds and covers a distance of 15m. Determine the acceleration of the jeepney ? Given: 0m/s……vi 3.25 s ….. T 15m……..d Required : acceleration (a)
solution d= vit + at 2 2 2d = vit + at2 a= 2d-vit t2 a= 2(15m) – 0 ( 3.25s) 2 a = 2.8 m/s 2
Assessment Solve the following problems: 1. With an initial velocity of 20 km/h, a car accelerated at 8 m/s 2 for 10 seconds. a) What is the position of the car at the end of the 10 seconds? b) What is the velocity of the car at the end of the 10 seconds? 2. A train accelerates to a speed of 20 m/s over a distance of 150 m. Determine the acceleration (assume uniform) of the train
Solve the following 3. : A car accelerates uniformly from 0 to 72 km/h in 11.5 seconds. a) What is the acceleration of the car in m/s 2 ? b) What is the position of the car by the time it reaches the velocity of 72 km/h?