Same force acts on two bodies of difference masses masses $2$ kg and $4$ kg initially at rest the ratio of times required to acquire same final velocity is.

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Solution

We know Force F=mass $\times$ acceleration= $a_{1} = \frac{F}{2}$
Similarly $a_{2} = \frac{F}{4}$
From equation of motion we get
$v_{1} = u + a_{1} t_{1}$
$v_{1} = \frac{F}{2} t_{1}$ -------(A)
$v_{2} = \frac{F}{4} t_{2}$ ------(B)
As per question $v_{1} = v_{2}$
$\frac{F}{2} t_{1} = \frac{F}{4} t_{2}$
$\frac{t_{1}}{t_{2}} = \frac{1}{2}$

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Same force acts on two bodies of difference masses masses 2 kg and 4 kg initially at rest the ratio of times required to acquire same final velocity is.

We know Force F=mass× acceleration=a1=F2Similarly a2=F4From equation of motion we get v1=u+a1t1v1=F2t1-------(A)v2=F4t2------(B)As per question v1=v2F2t1=F4t2t1t2=12

Same force acts on two bodies of difference masses masses $2$ kg and $4$ kg initially at rest the ratio of times required to acquire same final velocity is.