Unit 01 “Forces and the Laws of Motion” Problem Solving

Problem Solving Steps 1st List variables and assign values 2nd Choose an equation 3rd Plug In 4th Solve Equation 5th Box Answer

Easy Problem Solving Easy Problem Solving is 1-step. Fnet = F1+F2+ …. There are three equations to choose from. Fnet = F1+F2+ …. Net Force = Force 1 + Force 2 + … F=ma Force = mass x acceleration Fw=mg Weight = mass x gravity

1. Find the net force on a rope if it is pulled to the right with a force of 65N, pulled to the left with a force of 30N and left with a force of 20N.

1. Find the net force on a rope if it is pulled to the right with a force of 65N, pulled to the left with a force of 30N and left with a force of 20N. FNet=F1+F2+ F3 FNet= 65N + (-30N) + (-20N) FNet= 65N + (-50N) FNet= 15N Fnet = ? F1 = 65N F2 = -30N F3 = -20N

2. What is the net force on a skater who has a mass of 40kg and an acceleration of 2m/s2.

F = ma F = (40kg) x ( 2m/s2) F = 80 kgm/s2 F = 80 N F = ? m = 40kg 2. What is the net force on a skater who has a mass of 40kg and an acceleration of 2m/s2. F = ma F = (40kg) x ( 2m/s2) F = 80 kgm/s2 F = 80 N F = ? m = 40kg a = 2m/s2

3. What is the acceleration of a 30kg bicycle pushed with a force of 120N?

F = ma 120 N = (30kg) x (a) 120 kgm/s2 = (30 kg)(a) 30kg 30kg 3. What is the acceleration of a 30kg bicycle pushed with a force of 120N? F = ma 120 N = (30kg) x (a) 120 kgm/s2 = (30 kg)(a) 30kg 30kg 4m/s2 = a a = ? m = 30kg F = 120N

Complete the problems below on your whiteboards… A 160kg person is accelerated by an elevator at a rate of 6m/s2. What is the elevator’s force on the person? A box of books is pushed forward by two people, one with a force of 90N, the other with a force of 120N. IF friction pushed back on the box with a force of 75N, what is the net force on the box? What is the acceleration of a 50kg car if it is pushed forward with a force of 200N? (medium) What is the acceleration of a 50kg car if it is pushed forward with a force of 700N and pushed backwards with a force of 500N?

Medium Problem Solving Medium Problem Solving is 2-steps 1st: Use FNet equation 2nd: Use F=ma equation

4. A sailboat is pushed with the wind with a force of 160N North and also pushed North by the water current with a force of 110N and accelerates at a rate of 6m/s2. a) What is the net force on the boat? b) What is the mass of the boat?

F = ma 270 N = (m) x (6m/s2) 270 kgm/s2 = (m)(6m/s2) 6m/s2 6m/s2 4. A sailboat is pushed with the wind with a force of 160N North and also pushed North by the water current with a force of 110N and accelerates at a rate of 6m/s2. a) What is the net force on the boat? b) What is the mass of the boat? a) FNet=F1+F2 FNet= 160N + (110N) FNet= 270N Fnet = ? F1 = 160N F2 = 110N m = ? a = 6m/s2 F = 270N F = ma 270 N = (m) x (6m/s2) 270 kgm/s2 = (m)(6m/s2) 6m/s2 6m/s2 45 kg = m b)

5. What is the acceleration of a 3000kg airplane if the engine puts a force of 15000N forward and the air resistance applies a backwards force of 8000N?

F = ma 7000 N = (3000kg) x (a) 7000 kgm/s2 = (3000 kg)(a) 5. What is the acceleration of a 3000kg airplane if the engine puts a force of 15000N forward and the air resistance applies a backwards force of 8000N? FNet=F1+F2 FNet= 15000N + (-8000N) FNet= 7000N Fnet = ? F1 = 15000N F2 = -8000N F = ma 7000 N = (3000kg) x (a) 7000 kgm/s2 = (3000 kg)(a) 3000kg 3000kg 2.33m/s2 = a a = ? m = 3000kg F = 7000N

Complete the problems below on your whiteboards… A box of books is pushed forward by two people, one with a force of 50N, the other with a force of 275N. If the box has a mass of 40kg, what is the acceleration of the box? What is the mass of a car if it is pushed forward with a force of 9000N and pushed backwards with a force of 5500N and accelerates at a rate of 20m/s2? What is the mass of a boat if it accelerates at a rate of 12m/s2 under a water current of 300N East and wind pushing it with a force of 100N West? Completed all? Start your “pushing the learning” #28 Parts 4 and 5 on Problem Set 01

Hard and Difficult Problem Solving

Hard and Difficult Problem Solving Require understanding that forces have vector components. We learned that “net force” is the sum of the forces acting on an object. But we have dealt with objects with forces in only parallel directions (such as just in the x direction or just in the y direction). When forces are applied in both the X and Y direction you must use the Pythagorean theorem to find the resultant force of an X and Y force. or trigonometry to solve for the component forces, X and Y, of a net force.

Hard Problem Solving Fnet = √Fx2 + Fy2 θ = tan-1 (Fy/Fx) Requires two vectors to be combined into one resultant vector (Net Force) Fx = 94N Fy = 34N F = 100N Fnet Fnet = √Fx2 + Fy2 Fnet = √94N2 + 34N2 Fnet = √9992N2 Fnet = 100N Fy = 34N 20o Fx = 94N Angle θ = tan-1 (Fy/Fx) θ = tan-1(34N/94N) θ = 20o Fnet = √Fx2 + Fy2 θ = tan-1 (Fy/Fx)

HARD: The forces acting on a raft are 200N forward and 90N to the right. If the raft has a mass of 150kg, what are the magnitude and direction of the raft’s acceleration? Fx = 200N FY = 90N m = 150kg a = ? 24o 1st: Fnet Fnet = √Fx2 + Fy2 Fnet = √200N2 + 90N2 Fnet = √48100N2 Fnet = 219N Resultant Force Net Force 3rd: acceleration Fnet = ma 219N = (150kg)a 1.45m/s2 = a 2nd: Angle θ = tan-1 (Fy/Fx) θ = tan-1(90N/200N) θ = 24o a =1.45m/s2 24o below the horizontal

Difficult Problem Solving Requires a vector to be broken up into it’s X and Y components F = 100N θ= 20o Fx Fx = Fnetcosθ Fx = (100N)cos(20o) Fx = 94N F = 100N Fy = 34N 20o Fx = 94N Fy Fy = Fnetsinθ Fy = (100N)sin(20o) Fy = 34N Fx = Fnetcosθ Fy = Fnetsinθ

DIFFICULT: 30kg box is pulled by a worker with a force of 200N at an angle of 40o above the horizontal. What is the acceleration of the box? m = 30kg F = 200N θ= 40o a = ? F= 200N Fx 1st: Fx Fx = Fnetcosθ Fx = (200N)cos(40o) Fx = 153N 2nd: acceleration Fx = ma 153N = 30kg(a) 5.11m/s2 = a