What is the formula for work and how can it be defined?

[-Physician]

Homework Statement

We will start with the work formula $A=Fd$, can that formula be defined as:
$A=(ma)Δd=(m\frac{v}{t})vt=(m\frac{v}{t})v_0 t + \frac{1}{2} at^2$ ??

Homework Equations

$A=Fd$

The Attempt at a Solution

Yes, it can be defined.

I think the answer will be yes, but I will be sure.ALSO: Can we define the work by velocity , mass or any other..?
If so, what's the formula

 

[LawrenceC]

The equation is incorrect. ALWAYS check your units. When things are added, units must be homogeneous. 1/2at^2 has the units of length, not work.

 

[-Physician]

The equation is incorrect. ALWAYS check your units. When things are added, units must be homogeneous. 1/2at^2 has the units of length, not work.

Thanks, anyways:

ALSO: Can we define the work by velocity , mass or any other..?
If so, what's the formula

, can you answer?

 

[LawrenceC]

In certain cases you can equate work to kinetic energy. For instance if you pushed a block (initially stationary) on a frictionless horizontal surface with a constant force of F for a distance d, the following equation can be written:

F * d = 0.5 * m * v^2

where

F is force
d is distance
m is mass
v is velocity

Note that the units to the left and right of equals sign are the same. This always is the case.

 

[-Physician]

In certain cases you can equate work to kinetic energy. For instance if you pushed a block (initially stationary) on a frictionless horizontal surface with a constant force of F for a distance d, the following equation can be written:

F * d = 0.5 * m * v^2

where

F is force
d is distance
m is mass
v is velocity

Note that the units to the left and right of equals sign are the same. This always is the case.

thanks, but can I know why $0.5$ is put there and why is the velocity squared $v^2$? What's the effect of that?

 

[-Physician]

$v^2=v_0+2aχd$ would give us the equation $E_k=\frac{1}{2}mv^2=\frac{1}{2}aFt^2$ E_k is Kinetic Energy, Now I understand the 0.5, But What I don't understand is why did you multiply the force $F$ with distance $d$?

 

[sankalpmittal]

$v^2=v_0+2aχd$ would give us the equation $E_k=\frac{1}{2}mv^2=\frac{1}{2}aFt^2$ E_k is Kinetic Energy, Now I understand the 0.5, But What I don't understand is why did you multiply the force $F$ with distance $d$?

See derivation of kinetic energy formula : http://www.indiastudychannel.com/projects/1644-Derivation-Kinetic-energy.aspx

http://answers.yahoo.com/question/index?qid=20110530195816AAVPotW

Work is simply defined as mathematical product of force times displacement in direction of force or resolved in component of that force. What to say when you do work ? Push wall and it does not move. You do no work. That's the concept which implies meaning of the word "work" in physics.

By the way , your equation is also incorrect...
Correct will be to write :
v²=v₀²+2ad

 

[-Physician]

See derivation of kinetic energy formula : http://www.indiastudychannel.com/projects/1644-Derivation-Kinetic-energy.aspx

http://answers.yahoo.com/question/index?qid=20110530195816AAVPotW

Work is simply defined as mathematical product of force times displacement in direction of force or resolved in component of that force. What to say when you do work ? Push wall and it does not move. You do no work. That's the concept which implies meaning of the word "work" in physics.

By the way , your equation is also incorrect...
Correct will be to write :
v²=v₀²+2ad

thank you. got it now :)