Question

The same horizontal force acts on two blocks of masses 4kg and 2kg resting on a smooth horizontal surface. The ratio of the
times taken by them to cover equal distance is​

Answer 1

Given:

The same horizontal force acts on two blocks of masses 4kg and 2kg resting on a smooth horizontal surface.

To find:

Ratio of time taken by the objects to cover equal distance.

Calculation:

Acceleration of 4kg body = F/4 and

Acceleration of 2kg body = F/2.

Let equal distance covered be s ;

For 4kg body :

$s = u(t1) + \dfrac{1}{2} a {(t1)}^{2}$

$= > s = 0 + \dfrac{1}{2} a {(t1)}^{2}$

$= > s = \dfrac{1}{2} a {(t1)}^{2}$

$= > s = \dfrac{1}{2} \times ( \dfrac{F}{4}) \times {(t1)}^{2}$

$= > {(t1)}^{2} = \dfrac{8s}{F}$

$= > t1 = \sqrt{ \dfrac{8s}{F} }$

For 2kg body :

$s = u(t2) + \dfrac{1}{2} a {(t2)}^{2}$

$= > s = \dfrac{1}{2} a {(t2)}^{2}$

$= > s = \dfrac{1}{2} \times ( \dfrac{F}{2}) \times {(t2)}^{2}$

$= > t2 = \sqrt{ \dfrac{4s}{F} }$

Required ratio :

$\therefore \: t1 : t2 = \sqrt{8} : \sqrt{4}$

$= > \: t1 : t2 =2 \sqrt{2} : 2$

$= > \: t1 : t2 = \sqrt{2} : 1$

So, final answer is:

$\boxed{ \sf{ \bold{ \: t1 : t2 = \sqrt{2} : 1}}}$