Newton's Second Law and Motion: Finding Force

In this article, the topic is the relationship between Newton's Second Law and Motion, and the displacement of an object.

The method here will demonstrate how to find the force needed to move an object.

How do you find the force needed given time (t), final displacement (df), initial displacement (di), initial velocity (vi), and the mass of the object?

Once again, it is necessary to combine two principles in order to see the full relationship.

You will have to find the rate of acceleration (a) using the distance equation, then use that to find the force (F).

So, you want to use

F=ma

to find F, but you don't have the a. That means finding a using the distance information given and the distance equation.

The distance formula is:

df =  di  +  (vi )(t)+ 1/2(a)(t²)

So, we'll be doing two steps.

Step 1: 

Use the distance equation to find a:

df =  di  +  (vi )(t)+ 1/2(a)(t²)

Step 2:

Then use the calculated a with Newton's Second Law

F = ma

to find the force.

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EXAMPLES

Example 1

Let's take a look at another example (even easier!) and work it out:

What force acts on a object with a mass of 10 kg, if it begins 5 meters from a mark and has an initial velocity of 25 m/s, and ends up a final distance of 227 m from the mark after an elapsed time of 6 seconds?

STEP 1

Find a where:

df = 227 m
di = 5 m
vi = 25 m/s
t = 6 s

df =  di  +  (vi )(t)+ 1/2(a)(t²)
227 m = 5 m + (25 m/s)(6 s) + 1/2(a)(6²)

Combine some terms... and PEMDAS

227 m = 5 m + (25 m/s)(6 s) + 1/2(a)(6 s)²
227 m = 5 m + 150 m + 1/2(a)(36 s²)
227 m = 155 m + (18s²)(a)

Subtract 155 m from both sides...

78 m = (18s²)(a)

Divide both sides by 18s² ...

78 m / 18s² = a

4 m/s/s = a

Next, use THAT calculated a to find the Force (step 2 above):

Find F where

m = 10 kg

a = 4 m/s/s

F = ma
F = (10 kg )(4 m/s/s)
F = 40 N

Example 2

How about seeing one worked out?

https://youtu.be/MFxfo7jMg_M

NOTE
While this example shows how to find F when given distance information, the same process applies when asked to find the force given velocity information. However, you'd begin (Step 1) using the velocity information and formula to find a, the go on to Step 2 and solve for force.

SUMMARY:

You gotta do it in steps!

This process requires doing the work in steps. Depending on what is given, you use the two formulas below:

F = ma

df =  di  +  (vi )(t)+ 1/2(a)(t²)

So, read the problem, write down what is being looked for, write down what is given, THEN... use the two formulas. Two steps!

Newton's Second Law and Displacement

In this article, the topic is the relationship between Newton's Second Law and Motion, and this time, displacement will be the focus.

How do you find a total displacement when given time (t), initial displacement (di), initial velocity (vi), but instead of acceleration (a), you are given force (F) and mass (m)?

Once again, it is necessary to combine two principles in order to see the full relationship.

The distance formula is:

df =  di  +  (vi )(t)+ 1/2(a)(t²)

However, to find final displacement, you need an acceleration(a), but you have mass and force.

Using Newton's Second Law, fortunately, will allow the calculation of a using the force and mass given.

So, we'll be doing two steps.

Step 1: 

Use 

F=ma to find a

Step 2:

Then use THAT in the distance formula.

df =  di  +  (vi )(t)+ 1/2(a)(t²)

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EXAMPLES

Example 1

Let's take a look at another example (even easier!) and work it out:

A force of 50 N acts on a object with a mass of 10 kg. If it has an initial displacement (di) of 10 m and has an initial velocity (vi) of 20 m/s, what is its final displacement after an elapsed time of 4 seconds?

The question tells us that we need to find final velocity (vf). But there is no acceleration given. So, do step 1 (above):

Find a where:

F = 50 N

m = 10 kg

F = ma

50 N = (10 kg)(a)

50 N/10 kg = a

5 m/s/s = a

Next, use THAT calculated a to find final velocity (step 2 above):

Find df where

di = 10 m

vi = 20 m/s

t = 4 s

a = 5 m/s/s

df =  di  +  (vi )(t)+ 1/2 (a)(t²)

df = 10 m  +  (20 m/s )(4 s)+ 1/2 (5 m/s/s)(4²)

df = 10 m  +  80 m + 1/2 (5 m/s/s)(16 s²)

df = 10 m  +  80 m + 40 m
df = 130 m

Example 2

How about seeing one worked out?

https://youtu.be/TjxUEwMjmvc

SUMMARY:

You gotta do it in steps!

This process requires doing the work in steps. Depending on what is given, you use the two formulas below:

F = ma

df =  di  +  (vi )(t)+ 1/2(a)(t²)

So, read the problem, write down what is being looked for, write down what is given, THEN... use the two formulas. Two steps! 

Newton's Second Law and Final Velocity

Suppose you are faced with a problem such as:

A motorcycle stunt rider needs to reach a final velocity of 799 m/s in order to make the jump needed for the scene in the movie. If she has an initial velocity of 10 m/s and the total mass of her and the motorcycle is 125 kg and if she can create a force of 100 N, how much time will be needed to achieve the needed final velocity?

On the surface, it looks (exciting, but also) like this could be hard to do.

IT'S NOT!

The velocity formula is very easy:

vf = vi + (a)(t)

In the problem above, you are looking for t and vf and vi are given. And then there's that mass and force...

You need an acceleration(a), but you have mass and force. Thankfully Newton did that thing:

F = ma

So, we'll be doing two steps.

Step 1: 

Use 

F=ma to find a

Step 2:

Then use THAT in the velocity formula.

vf = vi + (a)(t)

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EXAMPLES

Example 1

Let's take a look at another example (even easier!) and work it out:

A force of 50 N acts on a object with a mass of 10 kg. If it has an initial velocity of 20 m/s, what is its final velocity after an elapsed time of 8 seconds?

The question tells us that we need to find final velocity (vf). But there is no acceleration given. So, do step 1 (above):

Find a where:

F = 50 N

m = 10 kg

F = ma

50 N = (10 kg)(a)

50 N/10 kg = a

5 m/s/s = a

Next, use THAT calculated a to find final velocity (step 2 above):

Find vf where

vi = 20 m/s

t = 8 s

a = 5 m/s/s

vf = vi + (a)(t)

vf = 20 m/s + (5 m/s/s)(8 s)

vf = 20 m/s + 40 m/s

vf = 60 m/s

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Example 2

How about seeing it worked out?

SUMMARY:

You gotta do it in steps!

This process requires doing the work in steps. Depending on what is given, you use the two formulas below:

F = ma

vf = vi + (a)(t)

So, read the problem, write down what is being looked for, write down what is given, THEN... use the two formulas. Two steps! 

Adhesion and Cohesion

What are adhesion and cohesion? To start with, they are properties of matter!

Cohesion is the tendency of molecules to stick to other molecules of the same substance. This is why drops of water bead up on a slick surface. In fact, this is why water forms drops in the first place!

The stronger the force of cohesion, the more molecules that will be able to stick to each other. Drops will be bigger!

Cohesion is also responsible for surface tension. If you've ever skipped a rock or ridden in a fast boat or jet ski, you have surface tension to thank!

Adhesion is the tendency of molecules to stick to other molecules of different substances. You probably already know what an adhesive does! An adhesive adheres to two things and makes them stick together. So, a good adhesive has a high degree of adhesion between itself and other things.

Consider this very fancy diagram (right)!

The GLUE adheres to the mug and it also adheres to the handle. Thus, the handle sticks to the glue which is stuck to the mug. The final result is that the handle sticks to the mug!

Adhesion exists beyond tapes and glues, too. The adhesive properties of water allow you to use it to stick the shower curtain to the tiles. If you have ever licked a plastic decal and stuck it to glass (or wiped it with water and stuck it to glass), it was the adhesive properties of water that allowed that to work.


Flash back to those beads of water… Car waxes brag that they cause water to bead up and run off.

http://carbondetail.com/

What's going on there? There is at play a combination of cohesion between the water molecules and adhesion between the water and the surface.

When water falls onto a surface, there is created a balance between the forces of attraction between the molecules to themselves and the surface. Wax makes the surface "slicker." That means that the adhesive force decreases, so the cohesive forces have a bigger effect. The water sticks to itself better than it does to the wax, so the beads of water become larger.

More Physical Properties of Matter

It has been established that…

A physical property of matter is any attribute, quality, or characteristic of a material that can be observed or measured without changing the composition of the substances in the material. 


Among the many properties of matter there are:

Viscosity
Conductivity
Malleability
Melting and Boiling Points
Density

These properties were introduced HERE.
http://billonscience.blogspot.com/2016/09/physical-properties-of-matter.html

There are many more physical properties of matter, and some of them will be introduced in this article.

Some physical properties are independent of the amount of the substance preset. Some physical properties change as the amount present changes.

If the property changes based on the how much is present, it is said to be an extensive property. Some factor outside the makeup of the material—some external factor is connected to the property. For instance, mass is an extensive property; the more of something you have, the more mass you have.

If the property does not change based on how much is present, it is said to be an intensive property. The property is independent of the amount present. For instance, color is an intensive property; no matter how much you have, the color is the same.

Physical Properties of Matter

Appearance (intensive)
How does it look? What is visually identifiable about the substance?

Some aspects of a physical appearance include its color, texture, or sheen. Uniformity of these things could also be a factor, or variances might indicate that the sample has some impurities in it.


Odor (intensive)

IMPORTANT NOTE: Don't sniff chemicals! You could die!

While it is dangerous to inhale chemicals, some of them do stimulate the olfactory nerves which results in the perception of odor.


Solubility (intensive)

Solubility is the degree to which a substance (solute) dissolves in a given solvent (other substance). Not everything dissolves in everything! There is a vast degree of variance in what will dissolve in what and to what degree! You need only watch a few TV commercials to know that Brand A dish soap will dissolve grease better than all the other! Let's just skip the laundry ads!

Solubility for a given substance might be noted in relationship to various solvents. How well does it dissolve in water? How about alcohol? How about mineral spirits? How about acetone?

Nail polish is a very intuitive example. Nail polish does not dissolve very well if at all in water. (If it did, it would come off with every hand wash!) It does, however dissolve readily in acetone (nail polish removers often are acetone).


Magnetivity / Magnetism (intensive*)

The degree to which a substance is attracted to or repelled by magnets and magnetic fields. Basically, do magnets stick to it or does it stick to iron?

*Larger samples of a magnetic substance will create a larger (i.e. stronger) magnetic pull, but the property is uniform regardless of sample size)


Ductility (intensive)

The degree to which a substance can elongate when pulled.

Examples: Chewing gum is very ductile. You can pull it and stretch it really far out of your mouth (although that is gross and germy). Carrots are not, compared to gum, very ductile. They snap off.


Specific Heat (intensive)

The capacity of a substance to hold energy in the form of heat.

If you have a dishwasher and have ever tried to unload it right after it was done, you know that the glass bowls will burn your fingers more than the plastic ones, but when the door opens, they are all the same temperature. Glass can hold heat better than plastic. Thus, the specific heat of glass is higher than the specific heat of plastic.

Specific heat is measured in a unit of energy per a unit of mass such as cal/gr or J/gr.


Opacity (intensive)

The capacity of a substance to block (usually limited to visible light) electromagnetic waves. Opacity ranges from transparent (light passes without diffusion), to translucent (some light passes but is diffused) to opaque (no light passes through).

While opacity is frequently used to discuss visible light, the same term applies to other types of radiation as well.


Mass (extensive)

The total amount of matter present, the sum of all the electrons, protons, and neutrons within the sample. 

Volume (extensive)

The total amount of space occupied by the sample.

Denisty (D) is an intensive property of matter that is the ratio of mass (m) to volume (V) found by:

D = m/V

Energy of Chemical Bonds - Basics

 General Chemistry Index

Where are we going with this? This page will give the ability to use laboratory observations and data to compare and contrast ionic, covalent, network, metallic, polar, and non-polar substances with respect to constituent particles, strength of bonds, melting and boiling points, and conductivity; provide examples of each type.

Energy of Chemical Bonds - Basics

Lead Author: Dr. Anne Gull

Chemical energy is a type of energy that is stored in the bonds of compounds. When compounds are formed, some energy is required to "shove" the atoms together. When those bonds are broken, that energy is released.

In typical chemical reaction, some bonds break (and give off energy) and other bonds form (taking in energy. If the total energy given off exceeds the energy need, the excess energy is given off as heat and/or light. When a reaction gives off energy, it is exothermic. Energy exits the reaction. When a reaction takes in energy, it is endothermic. Energy enters the reaction.

Every combination of atoms has a specific bond energy. Knowing the bond energy allows the calculation of energy given off.

For example, the bond between carbon and hydrogen stores 413,000 joules for every mole. That means that one mole of those bonds will give off that much energy.

Chemical energy is stored in bonds, so it’s a type of potential energy that can be released when a chemical reaction occurs. Each type of bond has a unique amount of energy. Some of the values are shown in the chart below. These energy amounts are measured for 1 mole.

The figure below shows the energy stored in several common bonds:

C-H	413000 J
C-C	347000 J
C-O	358000 J
C=O*	799000 J
O-H	467000 J
O-O	146000 J

* for CO2
For other C=O bonds, see this.

In order to figure out how much potential energy is inside a chemical substance, you need to look at the structure to see the bonds (the lines between elements represent the bonds). 

The dash or single line (-) is a single bond. The double line (=) is a different kind of bond, a double bond. There is also triple bonds (≡) three lines.

The energies for the different bonds are different. It is important to recognize the difference and make sure you are using the correct bond energy from the table.

More Bond Energies 

For more information and the energies of other types of bonds, see this link:

Click Here

https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Chemical_Bonding/Fundamentals_of_Chemical_Bonding/Bond_Energies


NOTE: The energies on the linked page are given in kilojoules, so multiply by 1000 to get joules per mole.

Example #1 Methane

https://www.daviddarling.info/encyclopedia/M/methane.html

In the molecule above, there are 4 C-H bonds, so the potential energy stored in this molecule is 4 times the value of each C-H (413000). So there is 4(413000) = 1652000 J of potential energy in one mole of this type of molecule.There are 3 C-H bonds, 1 C-O bond and 1 O-H bond so the potential energy stored in each bond would be added up.

Example #2 Methanol

https://www.daviddarling.info/encyclopedia/M/methanol.html

There are 3 C-H bonds, 1 C-O bond and 1 O-H bond so the potential energy stored in each bond would be added up.

3 C-H 3(413000) J

1 C-O 1(358000) J

1 O-H 1(467000) J

Total 2,064,000 J

Light and Optics: Introduction

Light

Light is one form of electromagnetic (EM) energy. Light is defined within a narrow range of electromagnetic frequencies which stimulate the receptors of the eye.  Included in light is the frequencies just beyond what humans can see. Infrared light has a slightly too low of a a frequency for humans to see. Ultraviolet light has a slightly too high frequency for humans to see.

All forms of electromagnetic energy are affected by interacting with matter. When light interacts with matter, it either reflects off of it,  refracts around it, or passes through.

One of the properties of light (and other EM types) is that it will propagate at different rates through different mediums. Though the difference is generally slight, it makes a difference.

Optics

Light, because it can be easily observed, provides many opportunities to EM radiation.

Notice how the pencil seems to bend. This occurs because
light travels at different rates through air and water (and glass).

The difference between the speed of light in a vacuum and the speed of light in air is very, very minimal (on a scale of X 10⁸⁾. However, the difference in speed between light in air and light in water water results in light bending at the waters surface.

The differences in speed also explain how lenses can reshape light. Glass can bend the light in one of two ways.

Focusing is the process of taking light that is moving in a straight line and concentrates it into a smaller area. The rays of light (particle model) are bent such that they converge (align into a tighter, smaller area).

Light can also be "un focused." Light that is traveling in a straight line can be dispersed into a wider area. The rays of light (particle model) are bent such that they diverge (spread out over a wider area).

Both lenses and mirrors can converge (focus) and diverge light. First… There are these three words you need to understand.

From Google

Concave: A shape that has a recessed center relative to the outer edges. The center caves in.

Flat: The center and the edges are in the same plane. (A window)

Convex: A shape that has a protruding center relative to the outer edges. NOT like a cave (see above).

So, which types of lens and mirror do what?

• Focus occurs with convex lenses and concave mirrors.
• Dispersion occurs with concave lenses and convex mirrors.

STOP - NOTICE!Throughout the vast domain of information on optics, lenses, and light, there exists a variety of sets of variables that identify different measures. The "distance to the object from the lens," for example can be found as… 

d… do… D… Do… u… o… 

This variation is very frequent. This discussion reflects that with examples and images used from other sources.

Focal Length

Every lens has a focal length. That is the distance from the lens to the point where parallel light will be focused:

An example of a convex mirror which spreads the light
making objects appear smaller than they actually are.

Convex lenses have a positive focal length. The image appears some distance behind the lens. The lens is between the object and the image.

Concave lenses have a negative focal length. No image appears, but the focal point is in front of the lens, between the object and the lens.

Concave mirrors have a positive focal length. The image appears some distance behind the lens. The lens is between the object and the image.

Convex mirrors have a negative focal length. No image appears, but the focal point is in front of the lens, between the object and the lens.

A flat mirror and a flat piece of glass (window) work into this schema, falling between convex and concave. Imagine a lens that is convex. As the radius of the curve increases, it gets closer and closer to a flat piece of glass (a window). Likewise, a mirror can flatten out as well.

The object appears inside the mirror an equal distance, but behind the mirror. More on this to come!

Depending on the substance from which a lens is made as well as the geometry of the lens, there is a relationship between the distance from the object to the lens, o, the focal length of the lens f, and the distance from the lens to the projected image i.

Object ------ o ------ Lens ----- i -------Image

That relationship is:

1/o + 1/i = 1/f

Solving for f… (consult with a math teacher if you have questions!)

(i/i)(1/o) + (o/o)(1/i) = 1/f

(i/io) + (o/io) = 1/f

(i + o)/(i)(o)  = 1/f


f = (i)(o)/(i + o)


So, knowing i and o, we can easily find f:

EXAMPLE:

An object is 2 cm from a lens and the image appears 5 cm from the lens. Find the focal length.

f = (i)(o)/(i + o)
f = (2 • 4)/(2 + 4)
f = 8 / 6
f = 1.333 cm

An interesting property to examine is the relationship of image size to object size in consideration of where the object is located and where the image appears.

Image from http://www.retremblay.net/PhyLifePart01/Individual_labs_files/lab10.pdf

The formula

Hi/Ho = Di/Do ( see eq. 1 above) 

is not difficult to work with. Cross multiplying will quickly reduce the problem to a simple solution, for instance:

EXAMPLE:

Find the height of the image, Hi, when the object has a height of 3 cm and where the object is 8 cm from the lens and the image is 4 cm from the lens.

Hi/Ho = Di/Do
Hi/3cm = 8cm/4cm 

cross multiply to get 

(4cm)(Hi) = 24cm² 

Hi = 6cm

Lastly, we can talk about magnification. How much bigger or smaller is the image compared to the object?.

This one is pretty easy!

M = hi/ho

So… divide…

EXAMPLE (without units, which would be a distance unit for hi and ho):

A lens has a magnification, M, of 1.25. I f the height of the object is 6, what is the height of the image?
M = hi/ho
1.25 = hi/6
(6)(1.25) = (hi/6)(6/1)
7.5 = hi

Conclusion

A very large portion of society has a daily encounter with optics: those who wear glasses or contacts and those take photos (camera or phone!)

While the need to calculate "things" rarely comes up, a fundamental understanding of lenses and light add to the daily life experience.

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¹ "propagates" is used to describe the motion of waves through a medium. Mechanical waves (such as sound) propagate by transferring energy through collision of molecules. Electromagnetic waves propagate differently and do not rely on matter as a medium. Once upon a time, scientists theorized that the universe was filled with something they called the aether that was responsible for the propagation of electromagnetic waves.