Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

Work

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?

share|improve this question

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – jinawee, Kyle Kanos, John Rennie, WetSavannaAnimal aka Rod Vance
If this question can be reworded to fit the rules in the help center, please edit the question.

2  
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 at 13:58
    
The work of my hand on the block? –  user37421 Jan 16 at 14:17
1  
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 at 14:23
    
OK that's great! –  user37421 Jan 16 at 19:28
add comment

4 Answers 4

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.

share|improve this answer
add comment

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.

share|improve this answer
add comment

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).

share|improve this answer
add comment

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.

share|improve this answer
1  
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 at 15:29
    
@rijulgupta Fixed, I hope :-) –  Carl Witthoft Jan 16 at 15:44
add comment

Not the answer you're looking for? Browse other questions tagged or ask your own question.