4.2 Kinetic energy and the work-energy theorem  (Page 2/6)

The $d$ cancels, and we rearrange this to obtain

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 $\frac{1}{2}{\text{mv}}^2$ . This quantity is our first example of a form of energy.

The quantity $\frac{1}{2}{\text{mv}}^2$ in the work-energy theorem is defined to be the translational kinetic energy    (KE) of a mass $m$ moving at a speed $v$ . In equation form, the translational kinetic energy,

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. The work-energy theorem can be compactly written as, $$W_{\mathrm{n}\mathrm{e}\mathrm{t}}=\mathrm{\Delta }\mathrm{K}\mathrm{E}$$ where ${\displaystyle \Delta }$ KE is understood to mean "change in KE."

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 40 m/s has four times the kinetic energy it has at 20 m/s, 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.

Determining the work to accelerate a package

Suppose that you push on the 30.0-kg package in [link] with a constant force of 120 N through a distance of 0.800 m, and that the opposing friction force averages 5.00 N.

(a) Calculate the net work done on the package. (b) Solve the same problem as in part (a), this time by finding the work done by each force that contributes to the net force.

Strategy and Concept for (a)

This is a motion in one dimension problem, because the downward force (from the weight of the package) and the normal force have equal magnitude and opposite direction, so that they cancel in calculating the net force, while the applied force, friction, and the displacement are all horizontal. (See [link] .) As expected, the net work is the net force times distance.