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 FnetFnet size 12{F rSub { size 8{"net"} } } {}. In equation form, this is Wnet=FnetdcosθWnet=Fnetdcosθ size 12{W rSub { size 8{"net"} } =F rSub { size 8{"net"} } d"cos"θ} {} where θθ size 12{θ} {} is the angle between the force vector and the displacement vector.

Figure 1(a) shows a graph of force versus displacement for the component of the force in the direction of the displacement—that is, an FcosθFcosθ size 12{F"cos"θ} {} vs. dd size 12{d} {} graph. In this case, FcosθFcosθ size 12{F"cos"θ} {} is constant. You can see that the area under the graph is FdcosθFdcosθ size 12{F"cos"θ} {}, or the work done. Figure 1(b) shows a more general process where the force varies. The area under the curve is divided into strips, each having an average force (Fcosθ)i(ave)(Fcosθ)i(ave) size 12{ \( F"cos"θ \) rSub { size 8{i \( "ave" \) } } } {}. The work done is (Fcosθ)i(ave)di(Fcosθ)i(ave)di 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 WiWi 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.

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 Figure 2.

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 FappFapp and the horizontal friction force ff. Thus, as expected, the net force is parallel to the displacement, so that θ=0ºθ=0º and cosθ=1cosθ=1 size 12{"cos"q=1} {}, and the net work is given by

Wnet=Fnetd.Wnet=Fnetd. size 12{W rSub { size 8{"net"} } =F rSub { size 8{"net"} } d} {}

(1)The effect of the net force FnetFnet size 12{F rSub { size 8{"net"} } } {} is to accelerate the package from v0v0 size 12{v rSub { size 8{0} } } {} to vv size 12{v} {}. The kinetic energy of the package increases, indicating that the net work done on the system is positive. (See Example 1.) By using Newton’s second law, and doing some algebra, we can reach an interesting conclusion. Substituting Fnet=maFnet=ma size 12{F rSub { size 8{"net"} } = ital "ma"} {} from Newton’s second law gives

Wnet=mad.Wnet=mad. size 12{W rSub { size 8{"net"} } = ital "mad"} {}

(2)To get a relationship between net work and the speed given to a system by the net force acting on it, we take d=x−x0d=x−x0 size 12{d=x - x rSub { size 8{0} } } {} and use the equation studied in Motion Equations for Constant Acceleration in One Dimension for the change in speed over a distance dd if the acceleration has the constant value
aa; namely,
v2=v02+2adv2=v02+2ad (note that
aa appears in the expression for the net work). Solving for acceleration gives
a=v2−v022da=v2−v022d. When
aa is substituted into the preceding expression for
WnetWnet, we obtain

Wnet=mv2−v022dd.Wnet=mv2−v022dd.

(3)The dd size 12{d} {} cancels, and we rearrange this to obtain

W
net
=
1
2
mv
2
−
1
2
mv
0
2
.
W
net
=
1
2
mv
2
−
1
2
mv
0
2
.
size 12{w"" lSub { size 8{ ital "net"} } = { {1} over {2} } ital "mv" rSup { size 8{2} } - { {1} over {2} } ital "mv""" lSub { size 8{0} } "" lSup { size 8{2} } "." } {}

(4)This expression is called the work-energy theorem, and it actually applies *in general* (even for forces that vary in direction and magnitude), although we have derived it for the special case of a constant force parallel to the displacement. The theorem implies that the net work on a system equals the change in the quantity 12mv212mv2 size 12{ { {1} over {2} } ital "mv" rSup { size 8{2} } } {}. This quantity is our first example of a form of energy.

The net work on a system equals the change in the quantity 12mv212mv2 size 12{ { { size 8{1} } over { size 8{2} } } ital "mv" rSup { size 8{2} } } {}.

W
net
=
1
2
mv
2
−
1
2
mv
0
2
W
net
=
1
2
mv
2
−
1
2
mv
0
2
size 12{w"" lSub { size 8{ ital "net"} } = { {1} over {2} } ital "mv" rSup { size 8{2} } - { {1} over {2} } ital "mv""" lSub { size 8{0} } "" lSup { size 8{2} } "." } {}

(5)
The quantity 12mv212mv2 size 12{ { {1} over {2} } ital "mv" rSup { size 8{2} } } {} in the work-energy theorem is defined to be the translational kinetic energy (KE) of a mass mm size 12{m} {} moving at a speed vv size 12{v} {}. (*Translational* kinetic energy is distinct from *rotational* kinetic energy, which is considered later.) In equation form, the translational kinetic energy,

KE
=
1
2
mv
2
,
KE
=
1
2
mv
2
,
size 12{"KE"= { {1} over {2} } ital "mv" rSup { size 8{2} } ,} {}

(6)is the energy associated with translational motion. Kinetic energy is a form of energy associated with the motion of a particle, single body, or system of objects moving together.

We are aware that it takes energy to get an object, like a car or the package in Figure 2, up to speed, but it may be a bit surprising that kinetic energy is proportional to speed squared. This proportionality means, for example, that a car traveling at 100 km/h has four times the kinetic energy it has at 50 km/h, helping to explain why high-speed collisions are so devastating. We will now consider a series of examples to illustrate various aspects of work and energy.

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