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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.

Applying the science practices: cars on a hill

Assemble a ramp suitable for rolling some toy cars up or down. Then plan a series of experiments to determine how the direction of a force relative to the velocity of an object alters the kinetic energy of the object. Note that gravity will be pointing down in all cases. What happens if you start the car at the top? How about at the bottom, with an initial velocity that is increasing? If your ramp is wide enough, what happens if you send the toy car straight across? Does varying the surface of the ramp change your results?

Sample Response: When the toy car is going down the ramp, with a component of gravity in the same direction, the kinetic energy increases. Sending the car up the ramp decreases the kinetic energy, as gravity is opposing the motion. Sending the car sideways should result in little to no change. If you have a surface that generates more friction than a smooth surface (carpet), note that the friction always opposed the motion, and hence decreases the kinetic energy.

Practice Key Terms 3

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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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