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What is the distance which the force is multiplied by it to get the work done?

a) The distance of the body motion while the force was touching it?

b) The total distance of the body motion?

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closed as off-topic by jinawee, Kyle Kanos, BMS, John Rennie, WetSavannaAnimal aka Rod Vance Apr 1 '14 at 8:37

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What work are you trying to find? The work of what on what? The work of the block on your hand? The work of your hand on the block? The work of the table on the block? The work of the block on the table? – NowIGetToLearnWhatAHeadIs Jan 16 '14 at 13:58
The work of my hand on the block? – Ahmed Elsawy Jan 16 '14 at 14:17
On questions like this you should explain what you have thought so far and try to narrow down what particular concept you are confused about. So for example look up the definition of work from a textbook or wikipedia article and try to apply it here. If you are confused about how to apply the definition, then ask a question about the specific thing that is confusing you, perhaps it is an ambiguity in the definition or something. – NowIGetToLearnWhatAHeadIs Jan 16 '14 at 14:23
OK that's great! – Ahmed Elsawy Jan 16 '14 at 19:28

Work, in particular this sort of mechanical work is essentially work done by a force on an object, it is defined as the dot(scalar) product of force with the distance the object moves due to the action of this force, in case of an infinitesimal distance, and in general an integration of this small work done over infinitesimal distance gives the required work done.

So to your particular question, it is the distance body moves under influence of force.

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If there is a difference between the distances in a and b, the force must have stopped and the book continued to move, coasting. In that case the force is zero for that time period. The integral of force*distance will then be the same, as the integral over the coast period will be zero.

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It is the distance of the body motion (anti)parallel to the direction of the force. So if force and velocity vector point in the same or in opposite directions, the work is defined as $W = |F| |s|$

If the force and velocity vectors are perpendicular, the work equals zero: the distance is zero.

If the force and velocity vectors are at an angle, then $W = |F| |s| \cos (\alpha)$, where $\alpha$ is the angle between force and velocity vector (needs to be integrated if the directions or magnitudes of either one of the vectors change).

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Rjul's and Ross's answers are correct but may be a bit confusing.

The typical Physics 101 textbook assumes that the work being done is just sufficient to overcome the restraining force such as gravity. As such, the object always stops dead the moment the force is removed. This lets one multiply the (presumed constant) force by the distance, a cheap integral.

Now, clearly :-) whether you fire a bullet vertically out of a cannon or raise the bullet on an elevator, the same net energy is required to achieve a given altitude (ignore air resistance and all the other nasty things that don't show up in introductory physics). Thus, in fact the integral of force over distance is what matters.

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I have edited my answer, actually I was trying to keep it to a very low level, I do infact know the importance of the integral. Anyways, I hope it's not inaccurate anymore, if it is, please tell me. – Rijul Gupta Jan 16 '14 at 15:29
@rijulgupta Fixed, I hope :-) – Carl Witthoft Jan 16 '14 at 15:44

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