Chapter 5. The Laws of Motion

Level : AP Physics Teacher : Kim

What causes changes in motion??

5.1 Concept of Force

- Forces cause change in the velocity of an object. So if an object moves with uniform motion(constant

speed), no force is required for the motion to be maintained.

- Forces are what cause objects to accelerate

- An object will accelerate only if the net force acting it is not equal to zero. That is, an unbalance force.

- If the net force exerted on an object is zero, then the acceleration of the object is zero and its velocity

remains constant

5.2 Newton’s 1st Law

In the absence of external forces, an object at rest remains at rest and an object in motion continues in

motion with a constant speed in a straight line

p.114 Quick Quiz5.1) True or False. (a) It is possible to have motion in the absence of a force. (b) It is

possible to have force in the absence of motion

5.3 Mass

- All objects have the tendency to resist changes in its state of motion. This is called inertia.

- Inertia and mass are directly related. That is, more mass means more inertia

- Mass and weight are two different quantities. => Weight = mass × gravitational acceleration

Q1) If an elephant was chasing after you, would it more advantage to you if you run away in a straight

line or zig-zag? Justify your answer~

Q2) If a book has a mass of 2kg, what is its weight?

5.4 Newton’s 2nd Law

- The acceleration of an object is directly proportional to the net force acting on it and inversely

proportional to its mass.

∑F = ma

where ∑Fx = max, ∑Fy = may, ∑Fz = maz

-The unit of force 1N = 1kg•m/s²

Q3) There is a 6kg block on a surface. The friction between the surface and the block is 3N. If you push

the object with a force of 10N, and the wind is blowing at the same direction of the motion with force of

5N, what is the net force? What is the acceleration of the object?

I. Applied Force(Fapp) at an Angle

Solve ‘Example 5.1’ in p.116 using Fx = max, ∑Fy = may

y

F2

x

F1

Q4) A crate with a mass of 23kg accelerates along a frictionless surface as the crate is pulled with

a force of 14.5N at an angle of θ=19°, as shown below. What is the horizontal acceleration of the

crate? ( See solution below for reference)

a) 1.4m/s2 b) 0.427m/s2 c) 1.29m/s2 d) 0.60m/s2

Fapp

θ

Solution for Q4)

step1) Draw an arrow to represent the force vector

step2) Draw x-y coordinate.

step3) Divide the force into components using trigonometry

Y FN

Fapp

Fapp sin19°

19° x

Fappcos19°

Fg =mg

step4) Using Newton's 2nd law to solve ; ∑F=ma

X : ∑Fx= Fappcos19°=max

Y : ∑Fy= FN +Fappsin19°–mg=may

We will assume that the block stays on the surface, so the block will only move in the x-direction and has

no vertical motion in the y-direction(ay=0). So we only need to use the equation in the x-direction.

Since Fapp=14.5N, solving for 'a' gives

ax= 14.5𝑐𝑜𝑠19°

23 =0.58m/s2

Q5) Two forces act on a 4.5kg block resting on a frictionless surface as shown below. What is the

horizontal acceleration of the block? F1 =5.9N, F2=3.7N and θ=43°

a) 1.8m/s2 b) 1.2m/s2 c) 0.82 m/s2 d) 0.14m/s2

F1

F2 θ

Q6) Two forces act on a 4.5kg block resting on a frictionless surface as shown below. What is the

horizontal acceleration of the block? F1 =5.9N, F2=3.7N and θ=43°

a) 1.8m/s2 , right b) 1.2m/s2 , right c) 0.82 m/s2 , right d) 0.14m/s2, right

F1

F2 θ

II. Motion on an Inclined Plane

- An inclined plane is one of the commonly-

recognized simple machines.

- Advantage

The inclined plane permits one to overcome a

large resistance by applying a relatively small

force through a longer distance than the load is to

be raised vertically

easier! harder!

Fapp Fapp

Q7) How much force is needed to give a 5kg block an acceleration of 2m/s² up the 30º incline

plane? Assume the inclined plane is frictionless. Solution is on next page!

Fapp

a) 45.6N b) 39.5N c) 34.5N d) 29.5N

Solution for Q7)

i) Draw a free-body diagram to identify the forces acting on the block

ii) Draw x-y coordinate. The x-axis must be parallel to the direction of

motion

y

FN Fapp x

Fg

iii) If any force is not aligned x or y axis, divide that force into

components

iv) Write the equations using ΣFx=max and ΣFy=may

ΣFx= Fapp - Fgsinθ = max , ΣFy= FN - Fgcosθ = 0 (since ay =0)

v) Solve for Fapp

Fapp = mgsinθ + max ( Fg = mg)

= 5×9.8×sin30° + 5×2 = 34.5N

y

FN

x

Fgsinθ Fapp

θ

Fgcosθ

Fg

Q8) Find the force needed to give a 5kg block an acceleration of 2m/s² up the incline plane when

the angle is 10° and 5°. Assume the inclined plane is frictionless.

Fapp

i) when θ=10°

ii) when θ=5°

Q9) An 8kg box is released on a 30º inclined plane and slides down a frictionless inclined plane.

i) Find the acceleration of the box. ii) Does the mass affect the motion of the box? (*~the box is

sliding due to gravity, so there is no applied force)

a) 4.9m/s2 b) 3.4m/s2 c) 2.2m/s2 d) 1.4m/s2

iii) From above, find the acceleration of the box when the angle is θ=20°, θ=10° and θ=5°.

Q10) A block is pushed up a frictionless 30° incline by an applied force as shown. If Fapp=25N and

m=3.0kg, what is the magnitude of the resulting acceleration of the block?

a) 2.3m/s2 b) 4.6m/s2 c) 3.5m/s2 d) 2.9m/s2 e) 5.1m/s

Fapp

30°

30°

5.5. The Force of Gravity and Weight

The weight of an object is defined as

Fg=mg

5.6 Newton’s 3rd Law

- If two objects interact, the force F12 exerted by object 1 on object 2 is equal in magnitude to and

opposite in direction to the force F21 exerted by object 2 on object 1:

F12 = –F21

- A force that affects the motion of an object must come from a second, external, object.

- A single isolated force cannot exist

Example) Cannonball Early naval cannons are allowed to roll backwards a little when fired.

Example 5.9 The Atwood Machine (Pulley System) see p.128 in the textbook When two objects of unequal mass are hung vertically over a frictionless pulley

of negligible mass as shown, the arrangement is called Atwood machine. The

device is sometimes used in the laboratory to determine the value of g.

Determine the magnitude of the acceleration of the two objects and the tension in

the lightweight string.

+

m1

m2

+

Normal Force

Q11) Any object with mass will be under the force of gravity toward the center of the Earth, if an object

is resting on a surface, why does the TV not accelerate in the direction of Fg?

Q12) A block of mass 4kg is resting on a surface.

i) Identify how many forces are acting on the block and draw a free-body diagram below that represents

the force.

ii) If 5N of force is being pushed down on the block, what is the normal force?

Ans) FN=44.2N

iii) If 10N of force is being pulled up by a rope, what is the normal force?

Ans) FN=29.2N

iv) If 5N of force is being applied horizontally(=Fapp) in the right direction, what is the normal force?

FN

Fapp

Fg

Ans) FN=39.2N

*v) If 5N of force is being pulled at an angle of 30° to the surface, what is the normal force?

FN

Fapp

30°

Ans) FN=36.7N

- Review ‘Conceptual Example 5.2’ in p.118

5.8 Forces of Friction ‘See video clip on coefficient of friction’

5.8.1Static friction & Kinetic friction

i) Static friction is friction between two solid objects that are not moving relative to each other. For

example, static friction can prevent an object from sliding down a sloped surface.

The static friction force must be overcome by an applied force before an object can move.

ii) Kinetic (or dynamic) friction occurs when two objects are moving relative to each other and rub

together (like a sled on the ground).

Kinetic friction is usually less than static friction.

5.8.2 Coefficient of Friction

- Coefficient of friction is ratio of the force of friction between two bodies and the force pressing them

together

- The coefficient of friction depends on the materials used; for example, ice on steel has a low coefficient

of friction, while rubber on pavement has a high coefficient of friction.

- Coefficients of friction range from near zero to greater than one – under good conditions, a tire on

concrete may have a coefficient of friction of 1.7.

Material Coefficient of Static Friction(μs) Coefficient of Kinetic Friction(μk)

Rubber on Concrete 1.0 0.8

Steel on Steel 0.74 0.57

Wood on Wood 0.3 0.2

Ice on Ice 0.1 0.03

5.8.3 Frictional Force

- What determines how much friction will act on the object?

=> The roughness of the surface(μ) and normal force FN

f = μFN

where f is the frictional force, μ is the coefficient of friction, FN is the normal force

Applying a Force on a Block at an Angle

Q13) A block of mass 10kg is resting on a flat surface. The coefficient of static friction μs=0.5.

i) What is the minimum horizontal force needed to move the object? See solution below!

ii) If the block is pulled at an angle of 30° from the surface. What is the force needed to move the object?

Ans) Fapp=43.9N

step1) Draw a free-body diagram representing all the forces

FN

f Fapp

Fg

step2) Use Newton’s 2nd law to set-up the equation

∑Fx = Fapp,x – f = max, ∑Fy = FN – Fg = may

- The moment the horizontal force(applied force) is equal to the frictional force, the object will begin to move

=> Fapp,x – f =0

- The vertical forces are equal since the block is in equilibrium state vertically

=> FN – Fg = 0

- Since f = μFN = mg, and Fapp,x - μFN = 0.

=> Fapp,x - μmg = 0

=> Fapp,x =μmg = 0.5×10×9.8 = 49N

∑F=ma, f = μ FN

Q14) A block of mass 20kg is resting on surface. The coefficient of kinetic friction μk=0.04. What is the

horizontal force needed to maintain constant speed?

Ans) Fapp=7.84N

Q15) An object of mass 5kg on a flat surface is pulled at an angle of 30° by a force of 6N. If the object is

moving at a constant speed, what is the coefficient of kinetic friction μk?

Ans) uk=0.113

Q16) Suppose that a 70kg box on a flat surface is pulled by a 400N force at an angle of 30° to the

horizontal. The coefficient of kinetic friction μk is 0.5. Find the acceleration of the box.(Note : the box is

not moving at a constant speed but accelerating!)

Ans) a=1.48m/s2

Q17) An 8kg box is released on a 30º incline plane and accelerates down the incline at 0.3m/s². Find the

frictional force impeding its motion. Also, what is the coefficient of kinetic friction μk in this situation?

Ans) f =36.8N, μk = 0.54

Q18) How much force is needed to give a 10kg block an acceleration of 0.2m/s² up the 40º incline if the

coefficient of kinetic friction μk is 0.3?

F=?

40°

Ans) Fapp=87.5N

Q19) A force of 50N is pushing a block of 5kg up an inclined plane of 30º above the horizontal. If μk is

0.3, what is the acceleration of the block?

F=50N

30°

Ans) a=2.55m/s2

6. Two-Block System

Tension Force

Forces are often applied by means of cables or ropes

that are used to pull an object, like a force T being

applied to the right of a rope attached to a block.

T

We always assume the rope is tight and strong so that there will be no compression or expansion

on the rope

Each particle in the rope, in turn, applies a force to its neighbor. As a result, the force is actually

directly applied to the block.

We also always assume the mass of the rope is very small compared to block so that we can

ignore it.

Q20) Without Friction A m1=10kg mass on a horizontal friction-free air track table is accelerated by a string attached

to another m2=10kg mass hanging vertically from a pulley. What is the acceleration of the

system of both masses? What is the tension force(T)? Solution in the next page

a) a=4.9m/s2, T=98N b) a=4.9m/s2, T=49N c) a=2.4m/s2, T=98N d) a=2.4m/s2, T=49N

m1

m2

*~Step by step solution~*

step1) Identify the forces by drawing a free-body diagram for both blocks separately

step2) Draw an x-y coordinate for both blocks.

step3) Use Newton’s 2nd law for both blocks separately: ΣF=ma

i) m1 X: ΣFx=T = m1a and Y : ΣFy=FN – Fg = 0 (since no vertical motion)

FN

T

Fg

ii) m2 X: no horizontal force exists Y : ΣFy= Fg –T = m2a where Fg =m2g

T

Fg

step4) Combine the equations for both masses and solve for ‘a’ and ‘T’

T = m1a

m2g – T = m2a

Q21) With Friction A m1=6kg block is on a horizontal surface. The coefficient of kinetic friction is µk=0.22. The

mass of the hanging block is m2=3kg. Find the acceleration a of both blocks and the tension T. m1

m2

a) a=2.55m/s2, T=18.2N b) a=1.83m/s2, T=23.9N

c) a=1.24m/s2, T=28.1N d) a=0.76m/s2, T=13.5N

∑F=ma, f = μ FN

Q22) The coefficient of kinetic friction between block m1 and the surface is µk=0.2. Also,

m1=25kg, m2=15kg. How far will block m2 drop in the first 3s after the system is released?

(*~use x‒xo=vot+½at2 to find the distance fallen)

m1

m2

d=?

a) 11m b) 14m c) 18m d) 29m

Q23) The coefficient of kinetic friction between block m1 and the table is µk=0.2. Also,

m1=25kg, m2=15kg. A horizontal force of Fapp=250N is pulling on the m1. What is the

acceleration of the system? The system is moving to the left

m1

Fapp

m2

a) 4.55m/s2 b) 3.49m/s2 c) 2.14m/s2 d) 1.35m/s2