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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

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

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

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

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 x 0 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 d if the acceleration has the constant value a ; namely, v 2 = v 0 2 + 2 ad (note that a appears in the expression for the net work). Solving for acceleration gives a = v 2 v 0 2 2 d . When a is substituted into the preceding expression for W net , we obtain

W net = m v 2 v 0 2 2 d d .

The d size 12{d} {} cancels, and we rearrange this to obtain

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} } "." } {}

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 1 2 mv 2 size 12{ { {1} over {2} } ital "mv" rSup { size 8{2} } } {} . This quantity is our first example of a form of energy.

The work-energy theorem

The net work on a system equals the change in the quantity 1 2 mv 2 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 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} } "." } {}

The quantity 1 2 mv 2 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 m size 12{m} {} moving at a speed v 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 , size 12{"KE"= { {1} over {2} } ital "mv" rSup { size 8{2} } ,} {}

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 [link] , 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.

Calculating the kinetic energy of a package

Suppose a 30.0-kg package on the roller belt conveyor system in [link] is moving at 0.500 m/s. What is its kinetic energy?

Strategy

Because the mass m and speed v are given, the kinetic energy can be calculated from its definition as given in the equation KE = 1 2 mv 2 size 12{"KE"= { {1} over {2} } ital "mv" rSup { size 8{2} } } {} .

Solution

The kinetic energy is given by

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

Entering known values gives

KE = 0 . 5 ( 30.0 kg ) ( 0.500 m/s ) 2 , size 12{"KE"=0 "." 5 \( "30" "." 0" kg" \) \( 0 "." "500"" m/s" \) rSup { size 8{2} } ,} {}

which yields

KE = 3.75 kg m 2 /s 2 = 3.75 J. size 12{"KE"=3 "." "75"`"kg" cdot m rSup { size 8{2} } "/s" rSup { size 8{2} } =3 "." "75"`J "." } {}

Discussion

Note that the unit of kinetic energy is the joule, the same as the unit of work, as mentioned when work was first defined. It is also interesting that, although this is a fairly massive package, its kinetic energy is not large at this relatively low speed. This fact is consistent with the observation that people can move packages like this without exhausting themselves.

Practice Key Terms 3

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Source:  OpenStax, College physics arranged for cpslo phys141. OpenStax CNX. Dec 23, 2014 Download for free at http://legacy.cnx.org/content/col11718/1.4
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