Name: Conservation of Mechanical Energy
Uploaded: 2019-07-19
Duration: 5 min 12 s
Description: When a body falls from a height towards the ground, its potential energy decreases but not the total mechanical energy. Let us understand this using data.

Question

Increase in

K E

of a car is

E_{1}

when it is accelerated from

15 m s^{- 1}

to

20 m s^{- 1}

and increase in

K E

of a car is

E_{2}

when it is accelerated from

20 m s^{- 1}

to

25 m s^{- 1} .

Find

\frac{E_{1}}{E_{2}}

.

Accepted Answer

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

$K . E = \frac{1}{2} m (v^{2} - u^{2})$ where, $m$ , $v$ and $u$ stands for Mass. Final Velocity and Initial Velocity respectively.

Increase in kinetic energy from $K E$ to $20 m / s$ i.e. $E_{1}$

$25 m s^{- 1} .$

$\Rightarrow E_{1} = \frac{1}{2} m [400 - 225]$

$\Rightarrow E_{1} = \frac{1}{2} m (175) . . . . . (1)$

Increase in kinetic energy from $20 m / s$ to $25 m / s$ ,

$\Rightarrow E_{2} = \frac{1}{2} m [{(25)}^{2} - {(20)}^{2}]$

$\Rightarrow E_{2} = \frac{1}{2} m [625 - 400]$

$\Rightarrow E_{2} = \frac{1}{2} m (225) . . . . . (2)$

Now, we are asked to find the ratio of $E_{1}$ and $E_{2}$

So, for that divide the equation $(1)$ by equation $(2)$

$\frac{E_{1}}{E_{2}} = \frac{[\frac{1}{2} m (175)]}{[\frac{1}{2} m (225)]}$

$\frac{E_{1}}{E_{2}} = \frac{175}{225}$

$\frac{E_{1}}{E_{2}} = \frac{7}{9}$

Therefore, required ratio is $7 : 9$ .

Increase in $K E$ of a car is $E_{1}$ when it is accelerated from $15 m s^{- 1}$ to $20 m s^{- 1}$ and increase in $K E$ of a car is $E_{2}$ when it is accelerated from $20 m s^{- 1}$ to $25 m s^{- 1} .$ Find $\frac{E_{1}}{E_{2}}$ .

Important Questions on Work, Energy and Power

Learn and Prepare for any exam you want !

Increase in KE of a car is E1 when it is accelerated from 15 m s-1 to 20 m s-1 and increase in KE of a car is E2 when it is accelerated from 20 m s-1 to 25 m s-1. Find E1E2.

Important Questions on Work, Energy and Power

Learn and Prepare for any exam you want !

Increase in $K E$ of a car is $E_{1}$ when it is accelerated from $15 m s^{- 1}$ to $20 m s^{- 1}$ and increase in $K E$ of a car is $E_{2}$ when it is accelerated from $20 m s^{- 1}$ to $25 m s^{- 1} .$ Find $\frac{E_{1}}{E_{2}}$ .