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The question is too vague: work done on what? Kinetic energy of what? If you mean: "is the work done on a body equivalent to the kinetic energy of that body?" then the answer is "in some cases it is, but in general it is not".

Generally, the work done by a force on a body equals the change in total mechanical energy of that body (this is called the work-energy theorem). The total mechanical energy of a body is equal to the sum of its kinetic energy and its potential energy. Potential energy is only relevant when the body is under the influence of what is called a conservative force, such as gravity.

We can express the work-energy theorem mathematically as follows:

KE1 + PE1 + W = KE2 + PE2 (Eq. 1)

Where W is the work done on the body under consideration, KE and PE represent kinetic energy and potential energy, respectively, of that body - and the subscripts 1 and 2 refer to the states just before and just after the work was done on the body.

If we picture a body that is far removed from any massive object (i.e., there is no significant gravity or any other conservative force field acting on the body), and we say furthermore that it is initially at rest with respect to our frame of reference, then we may set KE1, PE1, and PE2 equal to zero. Eq. 1 now becomes:

W = KE2

Or:

F d = 1/2 m v2

Where d is the distance over which force F acts, and m and v are the body's mass and velocity, respectively. Thus, in this case, work done does indeed equal the body's final kinetic energy. Remember, though, that this is only true in a special case, where simplifying assumptions have been made (no gravity, body initially at rest). Eq. 1, the general form of the work-energy theorem, generally holds in classical mechanics.

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16y ago

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