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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

A golfer on a fairway is 70 m away from the green, which sits below the level of the fairway by 20 m. If the golfer hits the ball at an angle of 40° with an initial speed of 20 m/s, how close to the green does she come?
Aislinn Reply
cm
tijani
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John Reply
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Siyaka Reply
A mouse of mass 200 g falls 100 m down a vertical mine shaft and lands at the bottom with a speed of 8.0 m/s. During its fall, how much work is done on the mouse by air resistance
Jude Reply
Can you compute that for me. Ty
Jude
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David Reply
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David
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emma Reply
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Youesf Reply
what is inorganic
emma
Chemistry is a branch of science that deals with the study of matter,it composition,it structure and the changes it undergoes
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Adjanou
chemistry could also be understood like the sexual attraction/repulsion of the male and female elements. the reaction varies depending on the energy differences of each given gender. + masculine -female.
Pedro
A ball is thrown straight up.it passes a 2.0m high window 7.50 m off the ground on it path up and takes 1.30 s to go past the window.what was the ball initial velocity
Krampah Reply
2. A sled plus passenger with total mass 50 kg is pulled 20 m across the snow (0.20) at constant velocity by a force directed 25° above the horizontal. Calculate (a) the work of the applied force, (b) the work of friction, and (c) the total work.
Sahid Reply
you have been hired as an espert witness in a court case involving an automobile accident. the accident involved car A of mass 1500kg which crashed into stationary car B of mass 1100kg. the driver of car A applied his brakes 15 m before he skidded and crashed into car B. after the collision, car A s
Samuel Reply
can someone explain to me, an ignorant high school student, why the trend of the graph doesn't follow the fact that the higher frequency a sound wave is, the more power it is, hence, making me think the phons output would follow this general trend?
Joseph Reply
Nevermind i just realied that the graph is the phons output for a person with normal hearing and not just the phons output of the sound waves power, I should read the entire thing next time
Joseph
Follow up question, does anyone know where I can find a graph that accuretly depicts the actual relative "power" output of sound over its frequency instead of just humans hearing
Joseph
"Generation of electrical energy from sound energy | IEEE Conference Publication | IEEE Xplore" ***ieeexplore.ieee.org/document/7150687?reload=true
Ryan
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Maurice Reply
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Maurice
answer
Magreth
progressive wave
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Mujahid
A string is 3.00 m long with a mass of 5.00 g. The string is held taut with a tension of 500.00 N applied to the string. A pulse is sent down the string. How long does it take the pulse to travel the 3.00 m of the string?
yasuo 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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