Question #47c4f

2 Answers
Jan 25, 2018

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

Explanation:

Have a look:
enter image source here

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%

Jan 25, 2018

"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%