I'm a little confused regarding the way Total work = Change in kinetic energy is derived using calculus. My issue can be seen at 3:26 of this video: https://youtu.be/2dqO4sy4Njg?t=3m20s

Why can the limits of the integral just be changed like that? how did it go from final/initial displacement to final/initial velocity?

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    $\begingroup$ See u-substitution $\endgroup$ – J. Murray Mar 18 '18 at 2:50
  • $\begingroup$ I'm familiar with u-substitution but can't figure out how to apply it for this proof. Could you post a derivation using u-substitution for clarification. $\endgroup$ – Cris Collante Mar 20 '18 at 8:47

Writing the work-KE theorem starting from (1D case for simplicity)

$$\int_{x_i}^{x_f} ma dx$$

is in principle wrong, for this already assume $a$ is a function of position.

The correct way should be

$$\int_{t_i}^{t_f}ma v dt=\int_{t_i}^{t_f}m\frac{dv}{dt}vdt=\int_{t_i}^{t_f}\frac{d}{dt}(\frac{1}{2}mv^2)dt=\frac{1}{2}mv(t_f)^2-\frac{1}{2}mv(t_i)^2$$

There is no need to do change of variable. Always stick to the independent variable $t$, which the integrand must be a function of. Otherwise, if you use $x$ or $v$, you have to worry whether the integrand is a function of them, which is a prerequisite that the change of variable is valid.

  • $\begingroup$ Sorry but could you clarify the final result a bit further. Shouldn't the final result give us a change in (1/2)mv^2. How would the independent variable t be used in calculations? $\endgroup$ – Cris Collante Mar 20 '18 at 8:50
  • $\begingroup$ By definition $\int_{t_i}^{t_f}\frac{d}{dt}f(t)dt=f(t_f)-f(t_i)$. $\endgroup$ – velut luna Mar 20 '18 at 8:52
  • $\begingroup$ @CrisCollante The final expression is not $t^2$. The velocity is written as a function of time, so $v(t_f)^2$ is the square of the velocity at the final time, which is the same as the velocity at the final position. $\endgroup$ – Mark H Mar 20 '18 at 22:05
  • $\begingroup$ Oh sorry I misunderstood, thanks for clearing that up Mark. I think I understand it now :) Thanks for the help! $\endgroup$ – Cris Collante Mar 21 '18 at 11:20

The change in limits is called substitution. Initially they assumed a to be a function of x. Therefore the work done was the change In position from x(i) to x(f) . But once they canceled dx the integral was a function of dv. In other words when x ranges from x(i) to x(f) v ranges from v(i) to v (f). This is just a shorter method u can in fact keep the limits as it is and integrate but once we integrate and this is just harder


d E = F dx

F = ma

a = dv/dt use the chain rule , given that dx/dt =v

a = [ dv/dt ] = [ dv/dx][dx/dt]

a = v[ dv/dx] , put all this information into dE

dE = F dx = ma dx = m [(v)(dv/dx)]dx , cancel dx then

dE = mv dv , now integrate and

E(1) - E(0) = [1/2] m v(1)^2 - [1/2]m v(0)^2


protected by Qmechanic Mar 20 '18 at 21:49

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