Calculating Percentage Change in Kinetic Energy with Increased Speed

[dimens]

Homework Statement

The kinetic energy K of a body of mass m is moving with speed v is given by..

K = 1/2mv^2

If a body's speed is increased by 2%, what is the approximate percentage change in it's kinetic energy...

The Attempt at a Solution

Δy ≈ f'(x) Δx

... f'(K) = mv Δx

mv* 2?

?

Don't know how to find the other values. Lol

 

[NascentOxygen]

Hi dimens. Let's write the equation: E= ½mv²

and we are asked to consider the case where (∆v)/v = 0.02

You have determined dE/dv = m⋅v

and we will approximate the actual ∆E/∆v by using dE/dv.

So far I have just restated what you provided. Can you complete this now?

Solution:
We use the equation: ∆E/∆v = ... http://smilearchive.com/s/cwm/3dlil/idea.gif

 

[dimens]

sorry i don't really get where you get the other variables from? E/0.02?

 

[NascentOxygen]

sorry i don't really get where you get the other variables from? E/0.02?

I just preferred E for K.E. because I thought E sounded more like an energy symbol!

As for where did I get that 0.02 from? I got it here:

If a body's speed is increased by 2%

 

[dimens]

but the answer is 4% i don't know where to get the other variables from :s

 

[NascentOxygen]

We don't need to get any more variables, we have all that is needed. The hard work has already been done. All that remains is some algebraic manipulation. So go back to my first post, follow through the steps as I've written them, and see whether you can complete the equation where I've indicated.