Question

Question #47c4f

Answer 1

I tried this...although I am not sure about tmy interpretation of the question...

Explanation:

Have a look:

So that the new Kinetic Energy is equal to the old one increased of an amount `n`:
`K_2=K_1+nK_1=K_1(1+n)`
where:
`1+n=2.25`
`n=2.25-1=1.25`
corresponding to `125%`

Answer 2

`"Details below."`

Explanation:

`"Let us rearrange the kinetic energy equation."`

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

`v=sqrt((2E)/m)`

`"let us multiply both sides of equation by m."`

`m*v=m*sqrt((2E)/m)`

`m*v=sqrt((2Ecancel(m^2))/cancel(m))`

`m v=sqrt (2mE)`

`"Let's write this equation twice for the initial and final cases."`

`"initial case:"`

`P_i=sqrt(2m E_i)" "(1)`

`"final case:"`

`P_i+0.5P_i=sqrt(2m(E_i+n E_i)`

`3/2P_i=sqrt(2m(E_i+n E_i))" "(2)`

`"let us find "((1))/((2))`

`cancel(P_i)/(3/2cancel( P_i))=sqrt((cancel(2m)E_i)/(cancel(2m)(E_i+n E_i)))`

`2/3=sqrt(E_i/(E_i+n E_i))`

`(2/3)^2=(sqrt(E_i/(E_i+n E_i)))^2`

`4/9=E_i/(E_i+n E_i)`

`9E_i=4 E_i+4nE_i`

`9E_i-4E_i=4nE_İ`

`5cancel(E_i)=4ncancel(E_i)`

`n=5/4" "125%`