Virtual Lab: Force, Distance, Work, and Energy

Important background information can be found here:

http://billonscience.blogspot.com/2017/04/concepts-of-force-work-and-energy.html

The lab will provide 3 trials in which force and distance are measured. Using the equation for work,

W = Fd

where W is work, F is applied force and d is distance through which the force was applied

it is relatively simple to calculate how much work was done.

Introductory science students should remember that when two variables are placed beside each other without any operator, it is meant for them to be multiplied.

The lab will provide an initial distance and a final distance that will be used to calculate distance.

The measures in the video are hypothetical, but reflect realistic quantities. They are contrived for the sake of creating a virtual lab experience that do not require actual lab access.

Instructions:

The virtual lab draws from the data in this video:

https://youtu.be/HbOrAmFhKH0



Watch the video and use the data within to complete the lab. The following link should open a COPY of a Google Doc into which you can type your answers.

CLICK HERE

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.

---

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)

---

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)

---

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

---

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