Physics and Work Help
Introduction to Physics and Work
In physics, work refers to a specific force applied over a specific distance. The most common examples are provided by lifting objects having significant mass (“weights” or “masses”) directly against the force of gravity. The amount of work w done by the application over a displacement q of a force whose magnitude is F is given by
w = Fq
The standard unit of work is the newton-meter (N · m), equivalent to a kilogram-meter squared per second squared (kg · m 2 /s 2 ).
Work As A Dot Product Of Vectors
The preceding formula is not quite complete because, as you should know by now, both force and displacement are vector quantities. How can we multiply two vectors? Fortunately, in this case it is easy because the force and displacement vectors generally point in the same direction when work is done. It turns out that the dot product provides the answer we need. Work is a scalar, therefore, and is equivalent to
w = F · q
where F is the force vector, represented as newtons in a certain direction, and q is the displacement vector, represented as meters in a certain direction. The directions of F and q are almost always the same. Note the dot symbol here (·), which is a heavy dot so that the dot product of vectors can be distinguished from the ordinary scalar product of variables, units, or numbers, as in kg · m 2 /s 2 .
As long as the force and displacement vectors point in the same direction, we can simply multiply their magnitudes and get a correct result for work done. Just remember that work is a scalar, not a vector.
Lifting An Object
Imagine a 1.0-kg object lifted upward against the Earth’s gravity. The easiest way to picture this is with a rope-and-pulley system. (Suppose that the pulley is frictionless and that the rope doesn’t stretch.) You stand on the floor, holding the rope, and pull downward. You must exert a certain force over a certain distance. The force and displacement vectors through which your hands move point in the same direction. You can wag your arms back and forth while you pull, but in practice this won’t make any difference in the amount of work required to lift the object a certain distance, so let’s keep things simple and suppose that you pull in a straight line.
The force of your pulling downward is translated to an equal force vector F upward on the object (Fig. 8-4 on p. 206). The object moves upward as far as you pull the rope, that is, by a distance q . What is the force with which you pull? It is the force required to exactly counteract the force of gravitation on the mass. The force of gravity F g on the object is the product of the object’s mass m and the acceleration vector a g of gravity. The value of a g is approximately 9.8 m/s 2 directly downward. To lift the object, you must exert a force F = m a g = (9.8 m/s 2 )(1.0 kg) = 9.8 kg · m/s 2 = 9.8 N directly upward.
Physics and Work Practice Problems
Fig. 8-4 . Work is done when a force is applied over a specific distance. In this case, the force is applied upward to an object against Earth’s gravity.
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