Understanding Work in Physics: Explained by the W=FD Formula

[cs23]

Mechanical Work is defined as W=FD. Where F is a constant force. The force is parallel to the displacement(providing theta = 0) and in the same direction.

Why do we multiply Force TIMES distance? It's the multiplication that confuses me

 

[Drakkith]

If you push a cart weighing 10 kg with a constant force, it takes double the work to move it twice the distance. Hence force x distance.

 

[russ_watters]

"Why" is a difficult question here. Ultimately, some guy saw FD as a useful thing to know and gave it the name "work". Beyond that, I'm really not sure what you are looking for.

 

[timthereaper]

Work is really about the transfer of energy (not heat, though) from one thing to another. The amount of energy imparted to an object is directly related to 1) the force acting on the object, and 2) the distance the object travels. Since the amount of energy imparted to an object increases when both force and distance increase, they get multiplied together.

 

[cs23]

Since the amount of energy imparted to an object increases when both force and distance increase, they get multiplied together.

Ok, so why not add the 2 instead?

 

[russ_watters]

Well for starters, you can't add quantities together that have different units...

 

[sophiecentaur]

Ok, so why not add the 2 instead?

Because the energy needed is proportional both to the force and two the distance. The X sign is appropriate for that sort of calculation - same as multiplying speed times time to get distance or multiplying cost per item times number of items to get the total cost.

 

[matphysik]

Mechanical Work is defined as W=FD. Where F is a constant force. The force is parallel to the displacement(providing theta = 0) and in the same direction.

Why do we multiply Force TIMES distance? It's the multiplication that confuses me

In 1D:

F·dx=ma·dx=mdv/dt·dx=mdv·dx/dt=mvdv⇒∫F·dx=½mv²+constant. Similarly for the `impulse`,
F·dt=ma·dt=mdv/dt·dt=mdv⇒∫F·dt=mv+constant.