Confused by the relationship of work and kinetic energy

  • #1
Oleiv
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So the equation for work is W = F * s
F = m * a, so W = m * a * s
Transferring this to units of measurement gives us: J = kg * m * s-2 * m
Or simplified: J = kg * m2 * s-2
Transferring back to units of quantity: W = m * v2
How can that be correct? Obviously Ekin = 1/2 * m * v2. Where did that 1/2 go? Or is W =/= Ekin? Am I making some other kind of mistake?
 

Answers and Replies

  • #2
sophiecentaur
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Where did that 1/2 go?
It's the rules of Integration. Integrate xdx and you get x2/2
Graphically, it's the area of the v/t triangle or the F/x triangle etc etc.
 
  • #3
PeroK
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So the equation for work is W = F * s
F = m * a, so W = m * a * s
Transferring this to units of measurement gives us: J = kg * m * s-2 * m
Or simplified: J = kg * m2 * s-2
Transferring back to units of quantity: W = m * v2
How can that be correct? Obviously Ekin = 1/2 * m * v2. Where did that 1/2 go? Or is W =/= Ekin? Am I making some other kind of mistake?

The ##\frac12## is effectively a conversion factor based on your choice of units. In any case, ##\frac12## is dimensionless and doesn't affect the dimensions or units in an equation.

For example, the area of a circle is ##A = \pi r^2##. Both quantities have dimensions of ##L^2## or SI units of ##m^2##. ##\pi## is a dimensionless, unit-less factor based on the geometry of the circle.
 
  • #4
DrClaude
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Transferring back to units of quantity: W = m * v2
How can that be correct? Obviously Ekin = 1/2 * m * v2. Where did that 1/2 go? Or is W =/= Ekin? Am I making some other kind of mistake?
To add a somewhat more physical explanation to the very good posts of @sophiecentaur and @PeroK, when you write the ##v## in the two equations is not the same. In ##E_\mathrm{kin}##, it is the final speed, when the work has stopped, while in the equation for work, the speed will change while work is applied. In the case of constant acceleration, we can write ##W = m \langle v \rangle^2 = mv_f^2/2##, with ##\langle v \rangle## the average speed while the work is applied.

Edit: this assumes also that the speed was initially 0. Otherwise, we would have to write the equations in terms of ##\Delta v## and ##\Delta E_\mathrm{kin}##.
 
  • #5
russ_watters
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So the equation for work is W = F * s
F = m * a, so W = m * a * s
Transferring this to units of measurement gives us: J = kg * m * s-2 * m
Or simplified: J = kg * m2 * s-2
Transferring back to units of quantity: W = m * v2
How can that be correct? Obviously Ekin = 1/2 * m * v2. Where did that 1/2 go? Or is W =/= Ekin? Am I making some other kind of mistake?
Dimension analysis is not the same as doing integration (or algebra if you assume constant acceleration). That's the mistake.

At constant acceleration, s=1/2at2

That's where the 1/2 comes from.
 

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