Science Starter! Draw a free-body diagram for: 1)A chair at rest on the floor. 2) A ball rolling to the right and slowing down across a grassy field.

1.) 2.)

according to Newton’s 1 st Law: - objects at rest remain at rest unless a force is applied to move them - objects in motion stay in motion unless a force is applied to change their speed or direction Remember that….

Newton’s Second Law The net force applied to an object to change its state of motion is directly proportional to the object’s mass and resulting acceleration of the object.

Newton’s 2 nd Law (formula form) F Net = ma F net (or ΣF): “Net Force” - The vector sum of all forces acting on an object [Newtons (N)] m: mass (kg) a: acceleration (m/s 2 )

Weight vs. Mass Weight: “F g ” force of gravity pulling on an object F = ma a  acceleration “g”  acceleration due to gravity (9.8 m/s 2 ) THEREFORE: F g = mg

Example A ball has a mass of 8.0 kg. What is the ball’s weight on earth ? (1) m = 8.0 kg g = 9.8 m/s 2 F g = ? (2) F g = mg (3) F g = (8.0 kg) (9.8 m/s 2 ) (4-5) F g = 78.4 N

Net Force Forces in “x” direction can be added, Forces in “y” direction can be added: Can be written as: F NET x OR Σ F x F NET y OR Σ F y CHEAT: Direction of acceleration is the POSITIVE DIRECTION

Example 1 A 10 kg wagon is being pulled to the right with 20 N of force. Between the tires and the ground, there is 12 N of friction. a) Draw a free-body diagram. b) Write an “F NET Equation” for the vertical and horizontal directions. c) Determine the acceleration of the wagon.

Σ F y = F N – F g Σ F x = F – F f (0) = F N – (10)(9.8)m(a) = F – F f 10 (a) = (20) – (12) 10 (a) = 8 a = 0.8 m/s 2

Two students are playing tug-of-war over a 15 kg crate. Joe pulls to the right with 40 N of force, while Bob pulls to the left with 60 N of force. The frictional force between the crate and the ground is 5 N. a) Draw a free-body diagram. b) Write an “F NET Equation” for the vertical and horizontal directions. c) Determine the acceleration of the crate. Example 2

Σ F y = F N – F g Σ F x = F Bob – F f – F Joe (0) = F N – (15)(9.8) m(a) = F Bob – F f – F Joe 15 (a) = (60) – (5) – (40) 15 (a) = 15 a = 1.0 m/s 2

Hooke’s Law: the Spring Force Robert Hooke: F s = kx F s : Spring Force (N) k: Spring Constant (N/m) x: Stretch distance (m)

Example 3 A 5 kg mass hangs from a spring of spring constant 120 N/m. How far will the spring stretch once the mass is at rest? a) Draw a free-body diagram of the mass. b) Write an “F NET Equation” for the vertical and horizontal directions. c) Determine the stretch distance of the spring. How many centimeters is this?

Σ F y = F s – F g Σ F x = 0 (m)(a) = (k)(x) – (m)(g) 5 (0) = (120)(x) – (5)(9.8) 0 = 120 (x) – = 120 x x = m = 40.8 cm