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  • Explain work as a transfer of energy and net work as the work done by the net force.
  • Explain and apply the work-energy theorem.

Work transfers energy

What happens to the work done on a system? Energy is transferred into the system, but in what form? Does it remain in the system or move on? The answers depend on the situation. For example, if the lawn mower in [link] (a) is pushed just hard enough to keep it going at a constant speed, then energy put into the mower by the person is removed continuously by friction, and eventually leaves the system in the form of heat transfer. In contrast, work done on the briefcase by the person carrying it up stairs in [link] (d) is stored in the briefcase-Earth system and can be recovered at any time, as shown in [link] (e). In fact, the building of the pyramids in ancient Egypt is an example of storing energy in a system by doing work on the system. Some of the energy imparted to the stone blocks in lifting them during construction of the pyramids remains in the stone-Earth system and has the potential to do work.

In this section we begin the study of various types of work and forms of energy. We will find that some types of work leave the energy of a system constant, for example, whereas others change the system in some way, such as making it move. We will also develop definitions of important forms of energy, such as the energy of motion.

Net work and the work-energy theorem

We know from the study of Newton’s laws in Dynamics: Force and Newton's Laws of Motion that net force causes acceleration. We will see in this section that work done by the net force gives a system energy of motion, and in the process we will also find an expression for the energy of motion.

Let us start by considering the total, or net, work done on a system. Net work is defined to be the sum of work done by all external forces—that is, net work    is the work done by the net external force F net size 12{F rSub { size 8{"net"} } } {} . In equation form, this is W net = F net d cos θ size 12{W rSub { size 8{"net"} } =F rSub { size 8{"net"} } d"cos"θ} {} where θ size 12{θ} {} is the angle between the force vector and the displacement vector.

[link] (a) shows a graph of force versus displacement for the component of the force in the direction of the displacement—that is, an F cos θ size 12{F"cos"θ} {} vs. d size 12{d} {} graph. In this case, F cos θ size 12{F"cos"θ} {} is constant. You can see that the area under the graph is F d cos θ size 12{F"cos"θ} {} , or the work done. [link] (b) shows a more general process where the force varies. The area under the curve is divided into strips, each having an average force ( F cos θ ) i ( ave ) size 12{ \( F"cos"θ \) rSub { size 8{i \( "ave" \) } } } {} . The work done is ( F cos θ ) i ( ave ) d i size 12{ \( F"cos"θ \) rSub { size 8{i \( "ave" \) } } d rSub { size 8{i} } } {} for each strip, and the total work done is the sum of the W i size 12{W rSub { size 8{i} } } {} . Thus the total work done is the total area under the curve, a useful property to which we shall refer later.

Two drawings labele a and b. (a) A graph of force component F cosine theta versus distance d. d is along the x axis and F cosine theta is along the y axis. A line of length d is drawn parallel to the horizontal axis for some value of F cosine theta. Area under this line in the graph is shaded and is equal to F cosine theta multiplied by d. F d cosine theta is equal to work W. (b) A graph of force component F cosine theta versus distance d. d is along the x axis and F cosine theta is along the y axis. There is an inclined line and the area under it is divided into many thin vertical strips of width d sub i. The area of one vertical stripe is equal to average value of F cosine theta times d sub i which equals to work W sub i.
(a) A graph of F cos θ vs. d size 12{d} {} , when F cos θ size 12{F"cos"θ} {} is constant. The area under the curve represents the work done by the force. (b) A graph of F cos θ size 12{F"cos"q} {} vs. d size 12{d} {} in which the force varies. The work done for each interval is the area of each strip; thus, the total area under the curve equals the total work done.

Net work will be simpler to examine if we consider a one-dimensional situation where a force is used to accelerate an object in a direction parallel to its initial velocity. Such a situation occurs for the package on the roller belt conveyor system shown in [link] .

Questions & Answers

what is math number
Tric Reply
x-2y+3z=-3 2x-y+z=7 -x+3y-z=6
Sidiki Reply
Need help solving this problem (2/7)^-2
Simone Reply
x+2y-z=7
Sidiki
what is the coefficient of -4×
Mehri Reply
-1
Shedrak
the operation * is x * y =x + y/ 1+(x × y) show if the operation is commutative if x × y is not equal to -1
Alfred Reply
An investment account was opened with an initial deposit of $9,600 and earns 7.4% interest, compounded continuously. How much will the account be worth after 15 years?
Kala Reply
lim x to infinity e^1-e^-1/log(1+x)
given eccentricity and a point find the equiation
Moses Reply
12, 17, 22.... 25th term
Alexandra Reply
12, 17, 22.... 25th term
Akash
College algebra is really hard?
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Carole
I'm 13 and I understand it great
AJ
I am 1 year old but I can do it! 1+1=2 proof very hard for me though.
Atone
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Adu
Not really they are just easy concepts which can be understood if you have great basics. I am 14 I understood them easily.
Vedant
find the 15th term of the geometric sequince whose first is 18 and last term of 387
Jerwin Reply
I know this work
salma
The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
virgelyn Reply
hmm well what is the answer
Abhi
If f(x) = x-2 then, f(3) when 5f(x+1) 5((3-2)+1) 5(1+1) 5(2) 10
Augustine
how do they get the third part x = (32)5/4
kinnecy Reply
make 5/4 into a mixed number, make that a decimal, and then multiply 32 by the decimal 5/4 turns out to be
AJ
how
Sheref
can someone help me with some logarithmic and exponential equations.
Jeffrey Reply
sure. what is your question?
ninjadapaul
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
ninjadapaul
I don't understand what the A with approx sign and the boxed x mean
ninjadapaul
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
ninjadapaul
oops. ignore that.
ninjadapaul
so you not have an equal sign anywhere in the original equation?
ninjadapaul
hmm
Abhi
is it a question of log
Abhi
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Abhi
I rally confuse this number And equations too I need exactly help
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salma
Commplementary angles
Idrissa Reply
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Sherica
im all ears I need to learn
Sherica
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Tamia
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Uday
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salma
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Ayuba
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opoku
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Ali
greetings from Iran
Ali
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Nharnhar
A soccer field is a rectangle 130 meters wide and 110 meters long. The coach asks players to run from one corner to the other corner diagonally across. What is that distance, to the nearest tenths place.
Kimberly Reply
Jeannette has $5 and $10 bills in her wallet. The number of fives is three more than six times the number of tens. Let t represent the number of tens. Write an expression for the number of fives.
August Reply
What is the expressiin for seven less than four times the number of nickels
Leonardo Reply
How do i figure this problem out.
how do you translate this in Algebraic Expressions
linda Reply
why surface tension is zero at critical temperature
Shanjida
I think if critical temperature denote high temperature then a liquid stats boils that time the water stats to evaporate so some moles of h2o to up and due to high temp the bonding break they have low density so it can be a reason
s.
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
Crystal Reply
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
Chris Reply
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Source:  OpenStax, Physics 110 at une. OpenStax CNX. Aug 29, 2013 Download for free at http://legacy.cnx.org/content/col11566/1.1
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