Work and Kinetic Energy

Work, symbolized W, is a change in energy produced by a force acting over a distance. When you lift a book off a table, you do work by exerting a force (to counteract gravity) over a distance (however high off the table you lift the book). Work is measured in joules, where [J]=[N m]=[kg m^2/s^2].

We define W=Int_c F dot ds for the work done by a force F acting over a path c. If the force is constant over the path, we can remove the force from the integral, leaving W=F dot Int_c ds. Since Int_c ds=delta s, W=F dot delta s iff the force is constant over the path.

Note that even though both of the inputs for the work function are vectors, they are combined through the dot product to make work a scalar quantity. Moreover, the dot product only takes into account parallel components, so in the previously-examined case of uniform circular motion, notice that no work is done by the centripetal force. The force is parallel to the acceleration—directed radially inwards—and each bit of displacement ds is perpendicular to the radius, leaving a dot product F dot ds=0. J.

Any object also has a defined kinetic energy K=(1/2)mv^2. Kinetic energy is also measured in joules. Since W=delta E, if the only change in energy of a system is in the system’s kinetic energy, then W=delta K=K_f-K_i, where K_f and K_i are the final and initial kinetic energies, respectively.

For example, if an object with m=5.00 kgis moving at v_i=10.0 m/s to the left and you exert a force that brings it to a stop over delta x=15.0 m, the initial kinetic energy is given by K=(1/2)(5.00 kg)(10.0 m/s)^2=250. J and the final kinetic energy is given by K=(1/2)(5.00 kg)(0. m/s)^2=0. J, we have W=delta K=-250. J. Since we know that delta x=15.0 m, we can find that the (average) force exerted by W=F dot delta s and plugging in -250. J=F dot (15.0 m), giving F=-16.7 J where the negative sign indicates that the force was exerted to the right.

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