Question #47c4f

2 Answers
Gió
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%#

ali ergin
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%#

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