Deriving kinetic energy from work using momentum

[Arman777]

I am trying to derive the kinetic energy from the work and can I derive it like this ?

$$W=\int Fdr$$
$$W=\int \frac {dp} {dt}dr=\int (dp) \frac {dr} {dt}=\int (mdv)v=1/2m[v_f^2-v_i^2]$$

 

[Arman777]

My book does this in 7 lines mine took 4. I don't know why books sometimes does things in long way.

 

[PeroK]

My book does this in 7 lines mine took 4. I don't know why books sometimes does things in long way.

Perhaps they prefer at least a modicum of mathematical rigour!

 

[Nugatory]

My book does this in 7 lines mine took 4. I don't know why books sometimes does things in long way.

There's a bit of personal taste involved here; what one person considers admirable terseness another may consider skipping important steps.

With textbooks an additional consideration is that skipping steps can be a problem for a student who isn't already familiar with the concept. For example, someone taking intro physics concurrently with their first calculus course may have seen their first integral just a few weeks back - it's easy to imagine that your cavalier treatment of the bounds of integration would confuse them.

For what it's worth... I understand your derivation just fine but it's not what I'd be writing on a chalkboard in front of a class.

 

[Arman777]

There's a bit of personal taste involved here; what one person considers admirable terseness another may consider skipping important steps.

With textbooks an additional consideration is that skipping steps can be a problem for a student who isn't already familiar with the concept. For example, someone taking intro physics concurrently with their first calculus course may have seen their first integral just a few weeks back - it's easy to imagine that your cavalier treatment of the bounds of integration would confuse them.

For what it's worth... I understand your derivation just fine but it's not what I'd be writing on a chalkboard in front of a class.

Well yes you are right. Thanks for your reply