step 8

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course phys 201

Question: `q001. Explain the difference between x - 2 / x + 4 and (x - 2) / (x + 4). Then evaluate each expression for x = 2.

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Your solution:

In the first problem the first step would be to divide 2 by x and then go in order, in the second the work in parentheses is done first.

2-2/2+4=

2-1+4=5

(2-2)/(2+4)=

0/6=0

confidence rating #$&*: 3

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Given Solution:

`aThe order of operations dictates that grouped expressions must be evaluated first, that exponentiation must be done before multiplication or division, which must be done before addition or subtraction.

It makes a big difference whether you subtract the 2 from the x or divide the -2 by 4 first. If there are no parentheses you have to divide before you subtract. Substituting 2 for x we get

2 - 2 / 2 + 4

= 2 - 1 + 4 (do multiplications and divisions before additions and subtractions)

= 5 (add and subtract in indicated order)

If there are parentheses you evaluate the grouped expressions first:

(x - 2) / (x + 4) = (2 - 2) / ( 2 + 4 ) = 0 / 6 = 0.

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Self-critique (if necessary):

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Self-critique Rating: 3

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Question: `q002. Explain the difference between 2 ^ x + 4 and 2 ^ (x + 4). Then evaluate each expression for x = 2.

Note that a ^ b means to raise a to the b power. This process is called exponentiation, and the ^ symbol is used on most calculators, and in most computer algebra systems, to represent exponentiation.

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Your solution:

In the first problem the power is one before addition, in the second the part in parentheses is done first.

2^2+4= 8

2^(2+4)=6

confidence rating #$&*: 3

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Given Solution:

`a2 ^ x + 4 indicates that you are to raise 2 to the x power before adding the 4.

2 ^ (x + 4) indicates that you are to first evaluate x + 4, then raise 2 to this power.

If x = 2, then

2 ^ x + 4 = 2 ^ 2 + 4 = 2 * 2 + 4 = 4 + 4 = 8.

and

2 ^ (x + 4) = 2 ^ (2 + 4) = 2 ^ 6 = 2*2*2*2*2*2 = 64.

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Self-critique (if necessary):

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Self-critique Rating:3

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Question: `q003. What is the numerator of the fraction in the expression x - 3 / [ (2x-5)^2 * 3x + 1 ] - 2 + 7x? What is the denominator? What do you get when you evaluate the expression for x = 2?

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Your solution:

Numerator: x-3

Denominator: everything else

Answer: -1/19

confidence rating #$&*: 3

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Given Solution:

`aThe numerator is 3. x isn't part of the fraction. / indicates division, which must always precede subtraction. Only the 3 is divided by [ (2x-5)^2 * 3x + 1 ] and only [ (2x-5)^2 * 3x + 1 ] divides 3.

If we mean (x - 3) / [ (2x-5)^2 * 3x + 1 ] - 2 + 7x we have to write it that way.

The preceding comments show that the denominator is [ (2x-5)^2 * 3x + 1 ]

Evaluating the expression for x = 2:

- 3 / [ (2 * 2 - 5)^2 * 3(2) + 1 ] - 2 + 7*2 =

2 - 3 / [ (4 - 5)^2 * 6 + 1 ] - 2 + 14 = evaluate in parenthese; do multiplications outside parentheses

2 - 3 / [ (-1)^2 * 6 + 1 ] -2 + 14 = add inside parentheses

2 - 3 / [ 1 * 6 + 1 ] - 2 + 14 = exponentiate in bracketed term;

2 - 3 / 7 - 2 + 14 = evaluate in brackets

13 4/7 or 95/7 or about 13.57 add and subtract in order.

The details of the calculation 2 - 3 / 7 - 2 + 14:

Since multiplication precedes addition or subtraction the 3/7 must be done first, making 3/7 a fraction. Changing the order of the terms we have

2 - 2 + 14 - 3 / 7 = 14 - 3/7 = 98/7 - 3/7 = 95/7.

COMMON STUDENT QUESTION: ok, I dont understand why x isnt part of the fraction? And I dont understand why only the brackets are divided by 3..why not the rest of the equation?

INSTRUCTOR RESPONSE: Different situations give us different algebraic expressions; the situation dictates the form of the expression.

If the above expression was was written otherwise it would be a completely different expression and most likely give you a different result when you substitute.

If we intended the numerator to be x - 3 then the expression would be written (x - 3) / [(2x-5)^2 * 3x + 1 ] - 2 + 7x, with the x - 3 grouped.

If we intended the numerator to be the entire expression after the / the expression would be written x - 3 / [(2x-5)^2 * 3x + 1 - 2 + 7x ].

STUDENT COMMENT: I wasn't sure if the numerator would be 3 or -3. or is the subtraction sign just that a sign in this case?

INSTRUCTOR RESPONSE: In this case you would regard the - sign as an operation to be performed between the value of x and the value of the fraction, rather than as part of the numerator. That is, you would regard x - 3 / [ (2x-5)^2 * 3x + 1 ] as a subtraction of the fraction 3 / [ (2x-5)^2 * 3x + 1 ] from the term x.

STUDENT QUESTION: There was another question I had about this problem that wasn’t addressed. At the end when you changed the order of operation from

2 - 2 + 14 - 3/7 = 14 - 3/7

where did the 98/7 - 3/7 come into play before the end solution of 95/7? I must have forgotten how to do this part.

INSTRUCTOR RESPONSE: It's not clear how you can get 95/7 without this step.

To do the subtraction 14 - 3/7 both terms must be expressed in terms of a common denominator. The most convenient common denominator is 7.

So 14 must be expressed with denominator 7. This is accomplished by multiplying 14 by 7 / 7, obtaining 14 * 7 / 7 = 98 / 7. Since 7/7 = 1, we have just multiplied 14 by 1. We chose to use 7 / 7 in order to give us the desired denominator 7.

Thus our subtraction is

14 - 3/7 =

98/7 - 3/7 =

(98 - 3) / 7 =

95 /7.

STUDENT COMMENT

It took me a while to think thru this one especially when I got to working with the fraction. Fractions have always been my

weak spot. Any tips to make working with fractions a little easier is greatly appreciated.

INSTRUCTOR RESPONSE

Fractions are seriously undertaught in our schools, so your comment is not unusual.

I have to focus my attention on the subject matter of my courses, and while I do address it to a point, I don't have time to do justice to the subject of fractions. In any case , to do so would be redundant on my part, since there are a lot of excellent resources on the Internet.

I suggest you search the Web using something like 'review of fractions', and find something appropriate to your needs. You should definitely review the topic, as should 95% of all students entering your course.

STUDENT COMMENT

I think I am confused on why the Numerator is not the top portion and denominator the bottom portion of the problem.

INSTRUCTOR RESPONSE

Everything is on one line so there is no top or bottom in the given expression. A numerator and denominator are determined by a division of two expressions.

As we know, a denominator divides a numerator. In the given expression the division sign occurs between the 3 and the [ (2x-5)^2 * 3x + 1 ], so 3 is the numerator and [ (2x-5)^2 * 3x + 1 ] is the denominator.

x is not divided by the denominator, since the division occurs before the subtraction. For the same reason the -2 + 7x is not involved in the division. So neither the x nor the -2 + 7 x is part of the fractional expression.

STUDENT COMMENT

Didn’t know that 3 / 7 was 3/7 as a

fraction.

INSTRUCTOR RESPONSE

3/7 is treated as a fraction because of the order of operations. 3 must be divided by 7 before any other operation is applied to either number, and 3 divided by 7 is the fraction 3/7.

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Self-critique (if necessary):

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Question: `q004. Explain, step by step, how you evaluate the expression (x - 5) ^ 2x-1 + 3 / x-2 for x = 4.

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Your solution:

confidence rating #$&*:

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Given Solution:

`aWe get

(4-5)^2 * 4 - 1 + 3 / 4 - 2

= (-1)^2 * 4 - 1 + 3 / 4 - 2 evaluating the term in parentheses

= 1 * 4 - 1 + 3 / 4 - 2 exponentiating (2 is the exponent, which is applied to -1 rather than multiplying the 2 by 4

= 4 - 1 + 3/4 - 2 noting that 3/4 is a fraction and adding and subtracting in order we get

= 1 3/4 = 7 /4 (Note that we could group the expression as 4 - 1 - 2 + 3/4 = 1 + 3/4 = 1 3/4 = 7/4).

COMMON ERROR:

(4 - 5) ^ 2*4 - 1 + 3 / 4 - 2 =

-1 ^ 2*4 - 1 + 3 / 4-2 =

-1 ^ 8 -1 + 3 / 4 - 2.

INSTRUCTOR COMMENTS:

There are two errors here. In the second step you can't multiply 2 * 4 because you have (-1)^2, which must be done first. Exponentiation precedes multiplication.

Also it isn't quite correct to write -1^2*4 at the beginning of the second step. If you were supposed to multiply 2 * 4 the expression would be (-1)^(2 * 4).

Note also that the -1 needs to be grouped because the entire expression (-1) is taken to the power. -1^8 would be -1 because you would raise 1 to the power 8 before applying the - sign, which is effectively a multiplication by -1.

STUDENT QUESTION: if it's read (-1)^8 it would be 1 or would you apply the sign afterward even if it is grouped and it be a -1?

INSTRUCTOR RESPONSE: The 8th power won't occur in this problem, of course, but you ask a good question.

-1^8 would require raising 1 to the 8th power, then applying the negative sign, and the result would be -1.

(-1)^8 would be the 8th power of -1, which as you see would be 1.

STUDENT COMMENT: I think it would be easier to visualize what your trying to raise to the exponent if you actually put parenthesis around the 2, that part seems to get tricky on the computer.

INSTRUCTOR RESPONSE: The expression was intentionally written to be misleading and make the point that, to avoid ambiguity, order of operations apply strictly, no matter what the expression looks like.

Normally, for clarity, the parentheses would be included. They aren't necessary, but when helpful it's a good idea to include them. You can, of course, have too many parentheses in an expression, making it harder than necessary to sort out. In practice we try to strike a balance.

The original expression was

(x - 5) ^ 2x-1 + 3 / x-2

White spaces make no difference in how an expression is evaluated, but they can help show the structure; e.g.,

(x - 5)^2 * x - 1 + 3 / x -2

is a visual improvement over the original. The * between the 2 and the x is not strictly necessary, but is also helpful.

((((x - 5) ^ 2)) * x) - 1 + (3 / x) - 2

verges on having too many parentheses at the beginning; it does help clarify the 3 / x.

STUDENT COMMENT

Although I read through your explanation and do see the point you are making, that 2x is actually 2 * x, I still think that

(-1) should be raised to 2x rather than 2. Kaking the answer -11/4, not 7/4.

INSTRUCTOR RESPONSE

When the expression (x - 5) ^ 2x-1 + 3 / x-2 is copied and pasted into a computer algebra system it is translated as

This notation is universal and unambiguous. Any deviation from strict interpretation (which does occur among some authors and among manufacturers of some calculators) tends to result in ambiguity and confusion.

STUDENT COMMENT

While I do understand what you are trying to relate, I will continue to make these mistakes on more than one occasion and will not penalize myself for not rewriting years of mathematics because of a syntax issue in an online class.

INSTRUCTOR RESPONSE

I don't penalize errors in typed notation when the intent is clear (though I will sometimes point out these errors), and when you take your tests you'll be writing them out by hand and this won't be an issue.

However this is not a syntax issue in an online class. This is the order of operations, as it has been since algebra was developed hundreds of years ago, and it's completely consistent with the mathematics you appear to know (quite well).

As stated here, if you use the wrong syntax in any computer algebra system, your expression will not be interpreted correctly. For this reason alone you need to understand the notation.

For this and other valid reasons you need to understand how the order of operations are represented in 'linear' fashion (i.e., 'typewriter notation') and to correctly interpret expressions written in this notatation.

Any mathematics that has been learned correctly is completely consistent with the order of operations and with the notation used in this course. If the mathematics you've learned was inconsistent with the order of operations (and I don't believe this is so in your case, but it is with many students), then you would need to adjust your thinking. Fortunately this is very easy to do. Interpret expressions literally, assume nothing, and everything works out.

You will also find that the notation quickly becomes easy to read and use, and that it expands your comprehension of all mathematical notation.

STUDENT COMMENT

I used -1^(2*4). I didn't realize that was doing multiplication before exponents. All of this typewriter notations seems ambiguous to me but I think that had I seen the expression in standard notation I would probably have made the same mistake in this instance. If I were writing this expression I would probably use a parenthesis or * to show the necessary separation.

INSTRUCTOR RESPONSE

Parentheses, even when they aren't strictly necessary, are often useful to clarify the expression. An parentheses, even when not necessary, are part of the order of operations.

Spacing is not part of the order of operations. An expression has the same meaning even if all spaces are removed.

However as long as an expression is correctly formed, spacing as well as parentheses can certainly be used to make it more readable.

I don't go to any trouble in this exercise to make the expressions readable, since my goal here is to make the point about order of operations, which give an expression its unambiguous meaning.

However in most of the documents you will be working with, I do make an effort to clarify the meanings of expressions through their formatting, often using unnecessary parentheses and spacing to help clarify meaning.

Certainly I encourage you to do the same.

STUDENT QUESTION

I didn’t separate the ¾ as a stand alone fraction, I am confused about why you don’t treat it as an equation that the

denominator isn’t treated as a denominator.

INSTRUCTOR RESPONSE

Your work was good throughout most of this problem. You did forget to copy down a -1 in one of the early steps, but otherwise followed the order of operations correctly until nearly the last step.

However near the end you said that 4+3/4-2=7/2.

You appear to have performed the addition 4 + 3 and the subtraction 4 - 2 before dividing. However the division has to be done first.

The division sign is between the 3 and the 4, so the division is 3/4, and that gives you the fraction 3/4.

Therefore the expression 4+3/4-2 tells you to 'add 3/4 to 4 then subtract 2'.

When actually writing this out we would probably include parentheses. That wasn't done here, as it would have defeated the point being made about order of operations, but for clarity we might have written

4 + (3/4) - 2.

The parentheses are not necessary around the 3/4, since the order of operations is sufficient to unambiguously define the result, but they do make the expression easier to read and reduce the likelihood of error.

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Self-critique (if necessary):

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Question: `q005. At the link

http://www.vhcc.edu/dsmith/genInfo/introductory problems/typewriter_notation_examples_with_links.htm

(copy this path into the Address box of your Internet browser; alternatively use the path

http://vhmthphy.vhcc.edu/ > General Information > Startup and Orientation (either scroll to bottom of page or click on Links to Supplemental Sites) > typewriter notation examples

and you will find a page containing a number of additional exercises and/or examples of typewriter notation.Locate this site, click on a few of the links, and describe what you see there.

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Your solution:

confidence rating #$&*:

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Given Solution:

`aYou should see a brief set of instructions and over 30 numbered examples. If you click on the word Picture you will see the standard-notation format of the expression. The link entitled Examples and Pictures, located in the initial instructions, shows all the examples and pictures without requiring you to click on the links. There is also a file which includes explanations.

The instructions include a note indicating that Liberal Arts Mathematics students don't need a deep understanding of the notation, Mth 173-4 and University Physics students need a very good understanding,

while students in other courses should understand the notation and should understand the more basic simplifications.

There is also a link to a page with pictures only, to provide the opportunity to translated standard notation into typewriter notation.

end program

STUDENT COMMENT (not quite correct)

I see a collection of typewriter problems, after looking at some of them I see that the slash mark is to create a fraction rather than to denote division.

INSTRUCTOR CORRECTION

A fraction is a division of the numerator by the denominator. The slash mark indicates division, which can often be denoted by a fraction.

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Self-critique (if necessary):

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Question: `q006 Standard mathematics notation is easier to look at; it's easier to see the meaning of the expressions.

However it's very important to understand order of operations, and students do get used to this way of doing it.

You should of course write everything out in standard notation when you work it on paper.

It is likely that you will at some point use a computer algebra system, and when you do you will probably have to enter expressions using a keyboard, so it is well worth the trouble to get used to this notation.

Indicate your understanding of why it is important to understand this notation.

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Your solution:

confidence rating #$&*:

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Self-critique (if necessary):

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Self-critique rating:

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Question: `q006 Standard mathematics notation is easier to look at; it's easier to see the meaning of the expressions.

However it's very important to understand order of operations, and students do get used to this way of doing it.

You should of course write everything out in standard notation when you work it on paper.

It is likely that you will at some point use a computer algebra system, and when you do you will probably have to enter expressions using a keyboard, so it is well worth the trouble to get used to this notation.

Indicate your understanding of why it is important to understand this notation.

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Your solution:

confidence rating #$&*:

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Self-critique (if necessary):

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Self-critique rating:

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@& You didn't complete this document.

Please complete all questions with answers and/or self-critiques and resubmit.*@

@& `q001. The meter, the kilogram and the second are the basic SI units (SI stands for 'standard international').

Velocity is the rate of change of position with respect to clock time. What therefore are the SI units of velocity?

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velocity is rate of change of position with respect to clock time = (change in position) / (change in clock time).

position has units of meters so change in position has units of m

clock time has units of seconds so change in clock time has unit of s

Therefore velocity has units of m / s.

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Acceleration is the rate of change of velocity with respect to clock time. What therefore are the SI units of acceleration?

****

acceleration is rate of change of velocity with respect to clock time = (change in vel) / (change in clock time).

change in vel has units of m/s

change in clock time has unit of s

so acceleration has units of (m/s)/s = m/s^2.

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A Newton is the SI unit of force. Force = mass * acceleration. In terms of kg, meters and seconds, what therefore are the units of of force? The resulting units are the units of a Newton.

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mass is in kg, acceleration in m/s^2, so force is in kg * m/s^2

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A Joule is the SI unit of work. Work = force * displacement. In terms of kg, meters and seconds, what therefore are the units of work? The resulting units are the units of a Joule.

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Force is measured in kg m/s^2, displacement in meters, so work is in kg m/s^2 * m = kg m^2 / s^2.

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Power is measured in watts. A watt is a Joule per second. What therefore are the fundamental units equivalent to a watt?

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work or energy is measured in kg m^2 / s^2 (i.e., Joules), time in s, so power is in units of Joules / sec = (kg m^2 / s^2) / s = kg m^2 / s^3.

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What are the fundamental units equivalent to Newtons * meters? What quantity would be expressed in these units?

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A Newton is a kg m/s^2, so N * m is kg m/s^2 * m = kg m^2 / s^2.

This was seen earlier to be the fundamental units of the Joule.

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What are the units of kinetic energy?

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ke=1/2kg*v^2

KE = 1/2 m v^2

m is in kg, v in m/s

So KE is in kg * (m/s)^2 = kg * m^2 / s^2.

Note that this was earlier shown to be equivalent to the Joule.

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`q002. Sketch an object on inclines of 10, 20 and 30 degrees; you will have three sketches. For each incline, sketch the arrow indicating the force exerted on the object by gravity (for each incline we'll refer to this as the 'first arrow'). For each sketch, construct the arrow which represents the component of the gravitational force parallel to the incline (you do this by projecting the arrow at a right angle onto a line parallel to the incline, as we did in class).

For each incline, estimate the length of this arrow as a percent of the first. Give your three estimated percents:

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An accurate sketch would show that the percents are about 17%, 34% and 50%.

Discrepancies could be due to errors in estimating the angle of the incline, projecting at an angle other than perpendicular to the line of the incline, or simply errors in estimating one vector as a percent of the other.

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`q003. My truck has a mass of about 1400 kg. It reaches the part of the road which has a nearly constant incline, in front of VHCC, at a speed of 5 meters / second in the direction down the incline. In 12 seconds its speed has increased to 10 meters / second.

What is my acceleration on that incline?

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acceleration is ave roc of vel wrt clock time = change in vel / change in clock time = (vf - v0) / `dt = (10m/s-5m/s)/12s=5m/s)/12s=.417m/s^2

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What is the net force on the truck as it coasts down that incline?

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F_net = m a = 1400kg*.417m/s^2=583.33kg*m/s^2.

Note that kg m/s^2 is the fundamental unit of the Newton, so the net force is about 580 N.

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What is the change in its kinetic energy as it coasts down the incline?

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KE_0 = 1/2 m v0^2 = 1/2 1400*(5m/s)^2= 700kg*25m^2/s^2=17500kg m^2/s^2

KE_f = 1/2 m vf^2 = 1/2 1400*(10m/s)^2= 700kg*100m^2/s^2=70000kg m^2/s^2

`dKE = KE_f - KE_0 = 70000kg m^2/s^2-17500kg m^2/s^2=52500kg m^2/s^2

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If friction exerts a force opposite to my direction of motion, with the magnitude of the frictional force equal to 2% of the force exerted by gravity on the truck, then during this interval how much work is done on the truck by the frictional force?

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The acceleration of gravity is 9.8 m/s^2 so the force exerted by gravity on the truck is

F_grav = m a_grav = 1400 kg * 9.8 m/s^2 = 13 700 kg m/s^2.

Note that this is the same as 13 700 Newtons.

Now, 2% of this gravitational force is about 270 Newtons.

The work by the frictional force can be denoted `dW_frict. Using this notation we conclude that `dW_frict = F_frict * `ds.

We need to find `ds for this motion.

The average velocity is (5 m/s + 10 m/s) / 2 = 7.5 m/s, and the interval lasts 12 seconds, so the truck moves 7.5 m/s * 12 s = 90 meters.

So F_frict * `ds = 270 N * 90 m = 2500 N * m = 2500 kg m^2 / s^2.

This is actually not quite right. The frictional force is in the direction opposite the displacement. The signs of F_frict and `ds are therefore opposite, and we conclude that the work done by friction is

`dW_frict = -2500 kg m^2 / s^2, or -2500 Joules.

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The net force on the truck is the combination of the frictional force, and the component of the gravitational force which acts in the direction down the incline. What therefore is that component of the gravitational force, and what percent is this of the total gravitational force pulling the truck toward the center of the Earth?

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The net force on the truck was found earlier to be about 580 Newtons; since the truck is speeding up this net force is in the direction of motion.

The frictional force was just found to be about 270 Newtons, and is in the direction opposite motion.

The net force is the sum of the frictional force and the 'parallel component' of the gravitational force, so

F_net = F_parallel + F_frict.

It follows that the parallel component of the gravitational force is F_net - F_frict.

Choosing the direction of motion as positive, we have

F_parallel = 580 Newtons - (-280 Newtons) = 580 N + 280 N = 860 N.

In fundamental units this is 860 kg m^2 / s^2.

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Make a reasonably accurate sketch depicting the incline, the truck, the gravitational force and its component along the incline.

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`q004. In the preceding, suppose that the direction down the incline is positive. The 5 meter / second initial velocity is therefore positive, and we would write v_0 = + 5 m/s. If the force of air resistance on the car was 2 kg m/s^2, then since that force is directed up the incline, it would be represented as -2 kg m/s^2.

Give each of the following, including its sign, its numerical value and its units:

The final velocity of the truck.

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The final velocity is in the direction down the incline so is positive.

Thus vf = +10 m/s.

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The change in the kinetic energy of the truck.

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As seen before, the change in KE is KE_f - KE_0. This was shown earlier to be about 50 000 kg m^2 / s^2, or 50 000 Joules.

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The acceleration of the truck.

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The acceleration is in the direction of motion, since the truck speeds up.

Thus, as found earlier, a = +.417m/s^2

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The force of friction on the truck.

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The force of friction was found earlier to have magnitude about 280 N.

This force is in the direction opposite motion.

So the frictional force is -280 N.

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The component of the gravitational force parallel to the direction of the incline.

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This was found earlier to be about 860 Newtons.

This force has to be greater than the net force, since friction contributes its negative share to the net force.

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The net force on the truck.

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F_net = m a = 1400 kg * (+.417 m/s^2) = +580 N, or +580 kg m/s^2.

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`q005. Suppose the truck coasts up the incline, starting at a velocity of 15 m/s, and continues until its velocity has decreased to 10 m/s. The frictional force still has a magnitude equal to 2% of the total gravitational force on the truck.

Let the direction down the incline be positive.

Give each of the following, including its sign, its numerical value and its units:

The final velocity of the truck.

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vf is still + 10 m/s.

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The change in the kinetic energy of the truck.

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KE_0 = 1/2 m v0^2 = 1/2 1400*(10m/s)^2=700kg*100m^2/s^2=70000kg*m^2/s^2

KE_f = 1/2 m vf^2 = 1/2 1400*(15m/s)^2=700kg*225m^2/s^2=157500kg*m^2/s^2

So

`dKE = KE_f - KE_0 = 70000kg*m^2/s^2-157500kg*m^2/s^2=-87500kg*m^2/s^2

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The force of friction on the truck.

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The force of friction is, as before, about -280 N.

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The component of the gravitational force parallel to the direction of the incline.

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This is as before about +860 N, or +860 kg m/s^2.

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The net force on the truck.

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The net force on the truck is unchanged, provided we continue to neglect air resistance.

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The acceleration of the truck.

****

The net force on the truck is unchanged so its acceleration is unchanged. Still + .417 m/s^2.

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`q006. This problem requires that you use the four equations of uniformly accelerated motion. If the truck reaches an incline on which its acceleration is .5 m/s^2, with velocity 10 m/s (both velocity and acceleration in the same direction), then how long will it take to reach a point 200 meters down the incline and how fast will it be moving at that point? You will need to carefully identify which of the quantities v0, vf, ds, dt and a are given. Then you should jot down the four equations of uniformly accelerated motion and select the one that most easily gives you additional information, and proceed from that point.

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.5 m/s^2 is the acceleration a.

10 m/s is its initial velocity v0.

200 meters is its displacement `ds.

If downward is the positive direction then all these quantities are positive.

Given a, v0 and `ds we can use the third or fourth equation to obtain additional information.

Using the fourth equation

vf^2 = v0^2 + 2 a `ds

we find that

vf = +- sqrt( v0^2 + 2 a `ds) = +- sqrt( (10 m/s)^2 + 2 * .5 m/s^2 * 200 m) = +- sqrt( 300 m^2 / s^2) = +- 17.3 m/s.

This velocity is downward, so we discard the negative solution and conclude that

vf = 17.3 m/s.

Its average velocity is therefore

vAve = (10 m/s + 17.3 m/s) / 2 = 13.7 m/s.

It travels the 200 meters in time interval `dt, such that vAve = `ds / `dt. Thus

`dt = `ds / vAve = 200 m / (13.7 m/s) = 15 sec, very roughly.

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`q007. This problem is fairly challenging. I expect all University Physics students to get it, and hope that at least some General College Physics students will also get it despite the fact that it's probably at least a little bit too challenging for this point of the course.

Suppose the truck is moving up the incline, on which its acceleration is .7 m/s^2 down the incline. It passes you moving at 12 meters / second, and then passes your friend, who is standing 16 meters up the incline from you.

How long does it take the truck to travel the intervening distance?

****

We can choose either up or down the incline as positive.

Let's choose up as positive.

Then the acceleration, being down the incline, is

-.7 m/s^2.

The car is initially moving up the incline so its initial velocity is

v0 = +12 m/s.

The displacement is also up the incline so

`ds = + 16 m.

Writing down the four equations of motion we find that we can most easily get additional information using the fourth equation

vf^2 = v0^2 + 2 a `ds

which yields

vf = +- sqrt( v0^2 + 2 a `ds) = +- sqrt( (12 m/s)^2 + 2 * (-.7 m/s) * 16 m) = +- sqrt(144 m^2 / s^2 - 22 m^2 / s^2) = +- sqrt(122 m^2 / s^2) = +-11 m/s, approximately.

The truck is still moving up the hill so

vf = + 11 m/s.

Its average velocity on the interval is

vAve = (vf + v0) / 2 = (12 m/s + 11 m/s) / 2 = 11.5 m/s, so the 16 meter displacement requires time

`dt = `ds / vAve = 16 m / (11.5 m/s) = 1.4 s, approx.

#$&*

How far up the incline does it go before coming to rest?

****

For the interval on which it comes to rest, starting at your position, we have vf = 0.

v0 is still 12 m/s and a is still -.7 m/s^2.

We can easily reason this out:

vAve = (vf + v0) / 2 = (12 m/s + 11.5 m/s) / 2 = 11.75 m/s.

`dv = vf - v0 = 0 - 12 m/s = - 12 m/s.

Since a = `dv / `dt, we have `dt = `dv / a = -12 m/s / (-.7 m/s^2) = 17 sec, approximately.

At average velocity 11.75 m/s, in 17 s the car travels about

`ds = vAve * `dt = 11.75 m/s * 17 s = 190 m.

#$&*

If it then coasts back down the incline, accelerating at .5 m/s^2, how long will it take to travel from the position of your friend to your position?

****

The car has 190 meters to travel back to you, starting from rest.

For the corresponding interval, still using up as positive, we have

`ds = -190 m

a = -.5 m/s^2

v0 = 0

so that

vf = +- sqrt( v0^2 - 2 a `ds) = ... = +- sqrt(190 m^2 / s^2) = +- 13.7 m/s, very approximately.

The car reaches you with a downward velocity, so

vf = - 13.7 m/s.

To analyze the motion between your friend's position and yours we set up a new interval. The initial event is the car reaching your friend, the final event is the car reaching you.

For this interval, still using upward as positive

vf = -13.7 m/s

a = -.5 m/s^2

`ds = -16 meters.

The fourth equation again works:

vf^2 = v0^2 + 2 a `ds

Solving for v0 we get

v0^2 = vf^2 - 2 a `ds

so that

v0 = +- sqrt(vf^2 - 2 a `ds ) = +- sqrt((13.7 m/2)^2 - 2 * (-.5 m/s^2) * (-16 m) ) = +- sqrt( 174 m^2 / s^2) = +- 13.3 m/s, approx..

This velocity is downward, so it's negative and

v0 = -13.3 m/s.

The car reaches your friend at 13.3 m/s in the downward direction, and you at 13.7 m/s in the same direction, therefore averaging 13.5 m/s for a displacement of 16 meters (both down the incline).

The time required is therefore

`dt = `ds / vAve = 16 m / (11.5 m/s) = 1.4 sec, approx.

However if calculated more accurately, there is a difference between the time required for the car to travel up the incline past you and your friend, and the time required to travel the same distance coming down.

*@

Pasted from

@& Check the appended document (directly above) to compare your results, most of which are good.

You did have a few errors and a few omissions, so let me know if you have questions.*@