Objectives Investigate the relationship between mass, force, and acceleration. State Newton’s second law and give examples to illustrate the law. Solve problems involving force, mass, and acceleration.
Warm-up Find the weight in Newtons for: 10 kg = _____ N 100 kg = _____ N 1 kg = _____ N
Everything in the world does one of two things:
Or……
Demo WDhvwhttp:// WDhvw
Exploring Newton’s Second Law Demonstration –What do you observe? –What can you measure? –What can you change? Design two separate data collection strategies to determine how two factors affect the acceleration of the system: the net force on the system and the total mass of the system.
Activity Computer Activity on Exploring Forces
Newton’s Second Law of Motion The relation between acceleration and force. Acceleration is proportional to force and inversely proportional to mass. Force is vector Acceleration is a vector “sigma” symbol means summation If there is only one force present, then you can leave it out.
Newton’s Second law F net = ma (N) = (kg) (m/s 2 ) Force problems always must use these units! If you prefer triangles “net” means “the total sum” or you can use the symbol Σ a F net m
Newton’s Second Law Force = Mass x Acceleration Force is measured in Newtons ACCELERATION of GRAVITY(Earth) = 9.8 m/s 2 Weight (force) = mass x gravity (Earth) Moon’s gravity is 1/6 of the Earth’s If you weigh 420 Newtons on earth, what will you weigh on the Moon? 70 Newtons If your mass is 41.5Kg on Earth what is your mass on the Moon?
Newton’s Second Law WEIGHT is a measure of the force of ________ on the mass of an object measured in __________ gravity Newtons
4-4 Newton’s Second Law of Motion Force is a vector, so is true along each coordinate axis. The unit of force in the SI system is the newton (N). Note that the pound is a unit of force, not of mass, and can therefore be equated to newtons but not to kilograms.
Newton’s Second Law One rock weighs 5 Newtons. The other rock weighs 0.5 Newtons. How much more force will be required to accelerate the first rock at the same rate as the second rock? Ten times as much
2 nd Law: F net = m a The acceleration an object undergoes is directly proportion to the net force acting on it. Mass is the constant of proportionality. For a given mass, if F net doubles, triples, etc. in size, so does a. For a given F net if m doubles, a is cut in half. F net and a are vectors; m is a scalar. F net and a always point in the same direction. The 1 st law is really a special case of the 2 nd law (if net force is zero, so is acceleration).
What is Net Force? When more than one force acts on a body, the net force (resultant force) is the vector combination of all the forces, i.e., the “net effect.” F1F1 F2F2 F3F3 F net
Net Force & the 2 nd Law For a while, we’ll only deal with forces that are horizontal or vertical. When forces act in the same line, we can just add or subtract their magnitudes to find the net force. 2 kg 15 N 32 N F net = 27 N to the right a = 13.5 m/s 2 10 N
Units F net = m a 1 N = 1 kg m/s 2 The SI unit of force is the Newton. A Newton is about a quarter pound. 1 lb = 4.45 N
Graph of F vs. a In the lab various known forces are applied—one at a time, to the same mass— and the corresponding accelerations are measured. The data are plotted. Since F and a are directly proportional, the relationship is linear. F a
Slope F a Since slope = rise / run = F / a, the slope is equal to the mass. Or, think of y = m x + b, like in algebra class. y corresponds to force, m to mass, x to acceleration, and b (the y-intercept) is zero. FF aa
W = mg Weight = mass acceleration due to gravity. This follows directly from F = m a. Weight is the force of gravity on a body. Near the surface of the Earth, g = 9.8 m/s 2.
Two Kinds of Mass Inertial mass: the net force on an object divided by its acceleration. m = F net / a Gravitational mass: Compare the gravitational attraction of an unknown mass to that of a known mass, usually with a balance. If it balances, the masses are equal. Balance ? m Einstein asserted that these two kinds of masses are equivalent.
Inertial Mass =Gravitational Mass
Video e38d-430b-bfb1-c505e628a2d4/science- nfl-football-newtons-second-law-motionhttp://science360.gov/obj/video/58e e38d-430b-bfb1-c505e628a2d4/science- nfl-football-newtons-second-law-motion
Forces & Kinematics 1.Find net force (by combining vectors). 2.Calculate acceleration (using 2 nd law). 3.Use kinematics equations: v f = v 0 + a t x = v 0 t + a t 2 v f 2 – v 0 2 = 2 a x To solve motion problems involving forces:
Newton 2, Slide 25 Example: Pushing a Box on Ice. A skater is pushing a heavy box (mass m = 100 kg) across a sheet of ice (horizontal & frictionless). He applies a force of 50 N toward the right. If the box starts at rest, what is its speed v after being pushed for a time t = 5 s ? d = ? F = 50 N v = ? m a = ? t = 5 s
Newton 2, Slide 26 Example: Pushing a Box on Ice... Start with F net = ma. –a = F net / m. –a = (50 N)/(100 kg) = 0.5 N/kg = (0.5 kg m/s 2 )/kg –a = 0.5 m/s 2 –Recall that v f = v i + at(from definition of a) –So v = (0.5 m/s 2 ) (5 s) –v = 2.5 m/s v = 2.5 m/s d = ? F = 50 N m a = 0.5 m/s 2 t = 5 s
Newton 2, Slide 27 Example: Pushing a Box on Ice... Now, what distance will the block travel during the 5 seconds? –d = ½ a t 2 –d = (0.5)(0.5m/s 2 )(5 s) 2 –d = 6.25 m d = ? F = 50 N v = 2.5 m/s m a = 0.5 m/s 2 t = 5 s
Sample Problem A 1. F net 2. a 3. v after 5 s 4. x after 5 s Goblin 400 N Ogre 1200 N Troll 850 N Treasure 300 kg = 50 N left = m/s 2 left = m/s left A troll and a goblin are fighting with a big, mean ogre over a treasure chest, initially at rest. Find: = 2.08 m left
Problem 1 What average force is required to stop an 2100-kg car in 12.0 s if the car is traveling at 95 km/h?
Problem 2 What average force is needed to accelerate a 7.00-gram pellet from rest to 125 m/s over a distance of m along the barrel of a rifle? Newton 2 Slide 30
#11 in book A particular race car can cover a quarter- mile track (402 m) in 6.40 s starting from a standstill. Assuming the acceleration is constant, how many “g’s” does the driver experience? If the combined mass of the driver and race car is 485 kg, what horizontal force must the road exert on the tires? Newton 2 Slide 31
Application of Newton’s 2 nd Law 34-e38d-430b-bfb1- c505e628a2d4/science-nfl-football- newtons-second-law-motionhttp://science360.gov/obj/video/58e e38d-430b-bfb1- c505e628a2d4/science-nfl-football- newtons-second-law-motion Other examples of applications of Newton’s 2 nd Law. Newton 2 Slide 32
Homework Chapter 4 Questions on page 97 #s 3, 4, and 13 Problems #s 1, 3, and 9
34 Closure If a loaded truck that can accelerate at 1 m/s 2 loses its load and has three-fourths of the original mass, what acceleration can it attain from the same driving force. Kahoot