Answered By Rohtaz S

Kinetic Energy of an object is written as  $E=\frac{1}{2}mv^2$E=12 mv² 

Given Kinetic energy for an object of mass (m) and velocity (v) =2J

 $E=\frac{1}{2}mv^2=2J$E=12 mv²=2J 

 $=>mv^2=2\times2J=4J$=>mv²=2×2J=4J 

Now, mass is doubled and velocity is tripled

Let’s define new mass as m₁₌ 2m and new velocity as v₁=3v

So, New Kinetic Energy(E₁)

E₁= $\frac{1}{2}\left(m1\right)\left(v1\right)^2=\frac{1}{2}\left(2\times m\right)\left(3\times v\right)^2=9\times mv^2$12 (m1)(v1)²=12 (2×m)(3×v)²=9×mv² 

As calculated above,  $mv^2=4J$mv²=4J 

Therefore E₁= $9\times4J=36J$9×4J=36J 

Answered By Majid B

 $m_2=2.m_1$m₂=‍2.m₁ 

 $v_2=3.v_1$v₂=3.v₁  

 $E_{k1}=\frac{1}{2}.m_1.v_1^2$E_k1=12 .m₁.v₁²     =>    $.m_1.v_1^2=2.E_{k1}=2\times2.0=4.0$.m₁.v₁²=2.E_k1=2×2.0=4.0 J

 $?E_{k2}=\frac{1}{2}.m_2.v_2^2=\frac{1}{2}.\left(2m_1\right).\left(3v_1\right)^2=m_1.\left(9v_1^2\right)=9.m_1.v_1^2$?‍E_k2=12 .m₂.v₂²=12 .(2m₁).(3v₁)²=m₁.(9v₁²)=9.m₁.v₁²

 $E_{k2}=9\times4.0=36.0$E_k2=9×4.0=36.0 J  

Answered By Scott J

_{$E1=\frac{1}{2}\left(m1\right)\left(v1\right)^2=2J$E1=12 (m1)(v1)²=2J} 

_{$E2=\frac{1}{2}\left(m2\right)\left(v2\right)^2$E2=12 (m2)(v2)²} 

_{$m2=2\left(m1\right),v2=3\left(v1\right)$m2=2(m1),v2=3(v1)} 

_{$\frac{E2}{E1}=\frac{\frac{1}{2}\left(m2\right)\left(v2\right)^2}{\frac{1}{2}\left(m1\right)\left(v1\right)^2}=\frac{E2}{2}J$E2E1 =12 (m2)(v2)²12 (m1)(v1)² =E22 J}  

_{$E2=\frac{2\left(\frac{1}{2}\right)2\left(m1\right)\left(3\left(v1\right)\right)^2}{\frac{1}{2}\left(m1\right)\left(v1\right)^2}=2\left(2\right)\left(3\right)^2=36J$E2=2(12 )2(m1)(3(v1))²12 (m1)(v1)² =2(2)(3)²=36J}