Question

Increase in KE of a car is E1 when it is accelerated from 15 ms-1 to 20 ms-1 and increase in KE of a car is E2, when it is accelerated from 20 ms- to 25 ms-'. Find e1/e2​

Answer 1

Change in Kinetic Energy/ Increase in Kinetic Energy is calculated using the following formula :-

$\\\bullet\quad\underline{\boxed{\sf K.E=\dfrac{1}{2}m(v^{2}-u^{2} ) }}\bigstar$

where, m, v and u stands for Mass. Final Velocity and Initial Velocity respectively.

Case 1:-

increase in kinetic energy from 15m/s to 20m/s i.e. E₁

$\\\quad\rightarrow\quad\sf E_{1}=\dfrac{1}{2}m\bigg[(20)^{2}-(15)^{2}\bigg]$

$\\\quad\rightarrow\quad\sf E_{1}=\dfrac{1}{2}m\bigg[400-225\bigg]$

$\\\quad\rightarrow\quad\sf E_{1}=\dfrac{1}{2}m(175) ---(i)$

Case 2:-

increase in kinetic energy from 20m/s to 25m/s

$\\\quad\rightarrow\quad\sf E_{2}=\dfrac{1}{2}m\bigg[(25)^{2}-(20)^{2}\bigg]$

$\\\quad\rightarrow\quad\sf E_{2}=\dfrac{1}{2}m\bigg[625-400\bigg]$

$\\\quad\rightarrow\quad\sf E_{2}=\dfrac{1}{2}m(225)---(ii)$

Now, we are asked to find the ratio of E₁ and E₂

So, for that divide the equation (i) by equation (ii)

$\\\quad\rightarrow\quad\sf \dfrac{E_{1}}{E_{2}}= \dfrac{\bigg[\dfrac{1}{2} m(175)\bigg]}{\bigg[\dfrac{1}{2}m(225)\bigg] }$

$\\\quad\rightarrow\quad\sf \dfrac{E_{1}}{E_{2}} = \dfrac{175}{225}$

$\\\quad\rightarrow\quad\boxed{ \boxed{\sf \dfrac{E_{1}}{E_{2}} = \dfrac{7}{9} }}\bigstar$

$\therefore\underline{\sf The\:required\:ratio\: is\:\bold{7:9}}$

_____________________