/sci/ - Science & Math

[Anonymous No.10807101]

Please, help a self learning oldfag some physics.

I understand that by doing (positive) Work on an object you are transferring energy to it. I also get that the rate at wich you are transferring that energy is Power.

When my brain stalls is at working out the meaning of W=F*d or P=F*v. What exactly does distance have to do in W=F*d? Why does a force do more work if it is applied over a longer distance?

[Anonymous No.10807119]

Work is defined as Fd. It's relationship with energy comes from the work energy theorem which says that the change of kinetic energy due to a force applied to the object is equal to the work. If you don't know calculus, you will probably not understand how this result is derived. Try and derive it for a constant force.

[Anonymous No.10807120]

>>10807101
The easiest way to think about this is from a definition point of view. Compare the units of force ([math]kg m /s^{2}[/math]) to energy ([math] kg m^2 / s^{2}[/math]) and it's obvious that multiplying by distance can get you an energy.

To expand on that, you can more generally define it with
[math] W = \int F dx [/math]
which in the case of constant force is just the Fx you get.

[Anonymous No.10808182]

>>10807101
Because the work in that context it's only appliance of movement. How much movement was applied. Photons don't move, photons wave.

[Anonymous No.10808335]

>>10807101
Look at vertical distance. F is constant close to the surface of the earth. If you double d, i. e. you push/carry something twice as high, you need to put in twice as much energy.

[Anonymous No.10808389]

>>10807101
[math] \displaystyle

\left \{
\begin{array}{l}
P~power \\
U~voltage \\
I~current \\
W~work,energy \\
t~time \\
F~force \\
s~distance \\
m~mass \\
a~acceleration \\
v~speed \\
\\
P=UI\\
P= \dfrac{W}{t}= \dfrac{F \cdot s}{t}= \dfrac{ma \cdot s}{t}
= \dfrac{m \dfrac{v}{t} \cdot s}{t}= \dfrac{m \dfrac{s/t}{t} \cdot s}{t}
= \dfrac{m \cdot s^2}{t^3}
\end{array}
\right.
\\\\
\left \{
\begin{array}{l}
W~watt \\
V~volt \\
A~ampere \\
J~joule \\
s~second \\
N~newton \\
m~meter \\
kg~kilogram \\
\\
W=V \cdot A \\
W= \dfrac{J}{s}= \dfrac{N\cdot m}{s}= \dfrac{kg \cdot \frac{m}{s^2} \cdot m}{s}
= \dfrac{kg \cdot m^2}{s^3}
\end{array}
\right.

[/math]

[Anonymous No.10808506]

>>10808389
based

[Anonymous No.10808728]

>>10807101
Well if the force was constant in dragging 2 boxes along a horizontal surface, but you move box A only 1 meter, and box B 5 meters. Which one intuitively do you think you did more work moving?