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

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

3. 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

4. 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.

5. 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

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

7. 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

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

9. 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 kg
4m/s2 = a
a = ?
m = 30kg
F = 120N

10. 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?

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

12. 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?

13. 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/s m/s2
45 kg = m b)

14. 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?

15. 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 kg
2.33m/s2 = a
a = ?
m = 3000kg
F = 7000N

16. 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

17. Hard and Difficult Problem Solving

18. 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.

19. 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)

20. 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

21. 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

22. 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