Unit 01 “Forces and the Laws of

Motion”Problem Solving

Problem Solving Steps

1st List variables and assign values

2nd Choose an equation3rd Plug In4th Solve Equation5th Box Answer

Easy Problem SolvingEasy Problem Solving is 1-step.

There are three equations to choose from.

Fnet = F1+F2+ ….Net Force = Force 1 + Force 2 + …

F=maForce = mass x acceleration

Fw=mgWeight = 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 = 65NF2 = -30NF3 = -20N

FNet=F1+F2+ F3

FNet= 65N + (-30N) + (-20N)

FNet= 65N + (-50N)

FNet= 15N

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

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

F = ?m = 40kga = 2m/s2

F = maF = (40kg) x ( 2m/s2)F = 80 kgm/s2

F = 80 N

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

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

a = ?m = 30kgF = 120N

F = ma120 N = (30kg) x (a)120 kgm/s2 = (30 kg)(a) 30kg 30kg 4m/s2 = a

Complete the problems below on your whiteboards…1. A 160kg person is accelerated by an elevator at a rate of

6m/s2. What is the elevator’s force on the person?

2. 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?

3. What is the acceleration of a 50kg car if it is pushed forward with a force of 200N?

4. (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?

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?

Fnet = ?F1 = 160NF2 = 110N

a) FNet=F1+F2

FNet= 160N + (110N)FNet= 270Nm = ?

a = 6m/s2

F = 270N

F = ma270 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?

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 = 15000NF2 = -8000N

FNet=F1+F2

FNet= 15000N + (-8000N)FNet= 7000N

a = ?m = 3000kgF = 7000N

F = ma7000 N = (3000kg) x (a)7000 kgm/s2 = (3000 kg)(a) 3000kg 3000kg

2.33m/s2 = a

Complete the problems below on your whiteboards…1. 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?

2. 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?

3. 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?

4. 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 SolvingRequires two vectors to be combined into one

resultant vector (Net Force)Fx = 94N

Fy = 34N

F = 100N

Fx = 94N

Fy = 34N

20o

Fnet

Fnet = √Fx2 + Fy

2

Fnet = √94N2 + 34N2

Fnet = √9992N2

Fnet = 100NAngle

θ = tan-1 (Fy/Fx)

θ = tan-1(34N/94N)

θ = 20o

Fnet = √Fx2 + Fy

2

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?

Resultant ForceNet Force

Fx = 200NFY = 90Nm = 150kga = ?

1st: Fnet

Fnet = √Fx2 + Fy

2

Fnet = √200N2 + 90N2

Fnet = √48100N2

Fnet = 219N2nd: Angleθ = tan-1 (Fy/Fx)θ = tan-1(90N/200N)θ = 24o

3rd: accelerationFnet = ma219N = (150kg)a1.45m/s2 = a

a =1.45m/s2 24o below the horizontal

24o

Difficult Problem SolvingRequires a vector to be broken up into it’s

X and Y components

Fx

Fx = FnetcosθFx = (100N)cos(20o)Fx = 94N

Fy

Fy = Fnetsinθ

Fy = (100N)sin(20o)

Fy = 34N

F = 100N θ= 20o F = 100N

Fx = 94N

Fy = 34N

20o

Fx = FnetcosθFy = 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 = 30kgF = 200Nθ= 40o

a = ?

1st: Fx

Fx = FnetcosθFx = (200N)cos(40o)Fx = 153N 2nd: acceleration

Fx = ma153N = 30kg(a)5.11m/s2 = a

Fx

F= 200N