0
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ 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
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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

$\endgroup$

protected by Qmechanic Mar 20 '18 at 21:49

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.