PHYSICS Newtons Laws Overview Review of Newtons Laws

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PHYSICS Newton’s Laws Overview

PHYSICS Newton’s Laws Overview

Review of Newton’s Laws of Motion st 1 nd 2 rd 3 Objects in

Review of Newton’s Laws of Motion st 1 nd 2 rd 3 Objects in motion stay in motion* and objects at rest stay at rest if there is zero net force (balanced) ΣF = m·a (the forces will be unbalanced) Every force has an equal and opposite force

Inertia n Depends on mass ¡ More mass more resistance ¡ Less mass less

Inertia n Depends on mass ¡ More mass more resistance ¡ Less mass less resistance

Demo: NFL Hits

Demo: NFL Hits

Equilibrium n Equilibrium: Net force is zero (ΣF = 0) ¡ ΣFx = 0

Equilibrium n Equilibrium: Net force is zero (ΣF = 0) ¡ ΣFx = 0 ¡ ΣFy = 0 FN FEngine FAir Fg

Equilibrium n n ΣF = 0 Newton’s First Law applies An object in equilibrium

Equilibrium n n ΣF = 0 Newton’s First Law applies An object in equilibrium can be: ¡ in motion (straight line/constant speed) ¡ at rest FN Ff Fair Fengine Fg

Terminal Velocity n n Once the forces of air resistance and gravity become balanced

Terminal Velocity n n Once the forces of air resistance and gravity become balanced equilibrium is reached No more acceleration

Newton’s Second Law n

Newton’s Second Law n

Newton’s Second Law n Equations: ¡ ¡ ¡ F = ma a = F/m

Newton’s Second Law n Equations: ¡ ¡ ¡ F = ma a = F/m m=F/a F = net force m = Mass a= Acceleration

Use one of the equations you just wrote down…

Use one of the equations you just wrote down…

Acceleration n

Acceleration n

Weight vs Mass n Weight Force Fg Fg = m ·g Mass: Amount of

Weight vs Mass n Weight Force Fg Fg = m ·g Mass: Amount of matter (does not change) Weight: Pull of gravity (changes)

Weight Force (Fg) g = 9. 8 m/s 2 g = 1. 6 m/s

Weight Force (Fg) g = 9. 8 m/s 2 g = 1. 6 m/s 2 g = 26 m/s 2 m = 50 kg Fg = 490 N ( 110 lb) m = 50 kg Fg = 80 N ( 18 lb) m = 50 kg Fg = 1300 N ( 292 lb)

In-Class Problem #1 A 2000 kg car has a push force of 5000 N

In-Class Problem #1 A 2000 kg car has a push force of 5000 N from its engine. If it experiences a friction force of 3000 N determine it’s (a) acceleration, (b) weight and (c) the normal force acting on it. a = 1 m/s 2 Fg = 19, 600 N FN = 19, 600 N

Review of Newton’s Laws of Motion ΣF = 0 ΣF ≠ 0 First Law

Review of Newton’s Laws of Motion ΣF = 0 ΣF ≠ 0 First Law Second Law a = 0 m/s 2 Accelerates at rest in motion* stays at rest stays in motion* depends on net force depends inversely on mass * Straight line/constant speed

Friction Force that resists motion due to imperfections in surfaces FRICTION MOTION

Friction Force that resists motion due to imperfections in surfaces FRICTION MOTION

Two Types 1. Static (rest): Keeps object from moving 2. Kinetic (moving): Slows moving

Two Types 1. Static (rest): Keeps object from moving 2. Kinetic (moving): Slows moving object

Friction Force Equation n μs (static) μk (kinetic)

Friction Force Equation n μs (static) μk (kinetic)

Coefficient of Friction Table

Coefficient of Friction Table

In-Class Problem #2 A 30 kg desk is at rest on the floor. It

In-Class Problem #2 A 30 kg desk is at rest on the floor. It takes 200 N of force to start it in motion. Determine the static coefficient of friction between the desk and the floor. μs = 0. 68

In-Class Problem #3 Once the desk in the previous problem is set in motion

In-Class Problem #3 Once the desk in the previous problem is set in motion the 200 N force continues to be applied. Determine the acceleration of the desk if the coefficient of kinetic friction is 0. 52. a = 1. 57 m/s 2