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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.

Real world connections: work and direction

Consider driving in a car. While moving, you have forward velocity and therefore kinetic energy. When you hit the brakes, they exert a force opposite to your direction of motion (acting through the wheels). The brakes do work on your car and reduce the kinetic energy. Similarly, when you accelerate, the engine (acting through the wheels) exerts a force in the direction of motion. The engine does work on your car, and increases the kinetic energy. Finally, if you go around a corner at a constant speed, you have the same kinetic energy both before and after the corner. The force exerted by the engine was perpendicular to the direction of motion, and therefore did no work and did not change the kinetic energy.

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] .

A package shown on a roller belt pushed with a force F towards the right shown by a vector F sub app equal to one hundred and twenty newtons. A vector w is in the downward direction starting from the bottom of the package and the reaction force N on the package is shown by the vector N pointing upwards at the bottom of the package. A frictional force vector of five point zero zero newtons acts on the package leftwards. The displacement d is shown by the vector pointing to the right with a value of zero point eight zero zero meters.
A package on a roller belt is pushed horizontally through a distance d .

The force of gravity and the normal force acting on the package are perpendicular to the displacement and do no work. Moreover, they are also equal in magnitude and opposite in direction so they cancel in calculating the net force. The net force arises solely from the horizontal applied force F app and the horizontal friction force f . Thus, as expected, the net force is parallel to the displacement, so that θ = and cos θ = 1 size 12{"cos"q=1} {} , and the net work is given by

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Source:  OpenStax, Work and energy. OpenStax CNX. Nov 09, 2015 Download for free at http://legacy.cnx.org/content/col11902/1.1
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