Energy recovered from a regenerative braking system

[sanderalan]

Homework Statement

Energy recovered from a regenerative braking system. Where the maximum regenerative braking is 75% at 1.5g from 100 km/h for 0.5 seconds

Relevant Equations

E = (1/2) * (m) * (delta v ^ 2)

Start velocity is 100 km/h, at 1.5g for 0.5 seconds the end velocity will be 73,52 km/h. So the energy can be calculated if the mass is present, but it is not given. Is there a way to calculate the regenerated energy with the given information?

 

[jbriggs444]

E = (1/2) * (m) * (delta v ^ 2)

Where did you get this equation?

 

[scottdave]

Maybe the OP meant delta (v^2).

 

[sanderalan]

Where did you get this equation?

https://engineering.stackexchange.c...:text=And V=rω=d,of energy is not significant.

 

[jbriggs444]

https://engineering.stackexchange.com/questions/24993/calculating-the-braking-energy-of-a-vehicle#:~:text=And V=rω=d,of energy is not significant.

The equation is wrong. So let's look at that source to see whether you are misinterpreting it.

As I suspected, that equation does not appear on the page you reference.

 

[sanderalan]

The equation is wrong. So let's look at that source to see whether you are misinterpreting it.

As I suspected, that equation does not appear on the page you reference.

What about
"The linear energy change is equal to:

Ke linear change =(1/2)mv1^2−(1/2)mv2^2"??

Please understand that I am working with limited information and this is what I found, Physics is not my expertise but I want to understand what I am missing in this problem...

 

[berkeman]

Homework Statement:: Energy recovered from a regenerative braking system. Where the maximum regenerative braking is 75% at 1.5g from 100 km/h for 0.5 seconds
Relevant Equations:: E = (1/2) * (m) * (delta v ^ 2)

So the energy can be calculated if the mass is present, but it is not given. Is there a way to calculate the regenerated energy with the given information?

The KE is definitely dependent on the mass. Is there more to the question?

 

[jbriggs444]

What about
"The linear energy change is equal to:

Ke linear change =(1/2)mv1^2−(1/2)mv2^2"??

as @scottdave has noted, ${v_1}^2 - {v_2}^2$ is different from $(v_1-v_2)^2$.

You were right to notice that kinetic energy depends on mass and that knowing the initial and final velocities does not help determine how much energy is gained or lost unless you also know the mass.

 

[sanderalan]

Thanks! Seems like the question was not complete in this case, as no mass was given

 

[haruspex]

Thanks! Seems like the question was not complete in this case, as no mass was given

Depending on the way in which you are required to submit your answer, you could say "if the mass is m then the energy recouped is. ...m."