Question

If the extension of the spring is $x_0$ at time t, then the displacement of the second block at this instant is :

• A. $\frac{F{t}^2}{2m}-x_0$
• B. $\frac{1}{2}(\frac{F{t}^2}{2m}+x_0)$
• C. $\frac{1}{2}(\frac{2F{t}^2}{m}-x_0)$
• D. $\frac{1}{2}(\frac{F{t}^2}{2m}-x_0)$

Answer 1

The correct option is D $\frac{1}{2}(\frac{F{t}^2}{2m}-x_0)$

When a constant force $F$ is applied on the system , it is accelerated and the acceleration of the system is given by ,

$a=F/(m+m)=F/2m$ ,

now the displacement of the system can be given by ,

$d=ut+1/2a{t}^2$ ,

here , $u=0$ , as the system is at rest initially ,

therefore displacement of the system (center of mass of the system) in time $t$ will be ,

$d=0+1/2a{t}^2=\frac{F{t}^2}{4m}$ ,

as two masses are equal therefore center of mass of the system will be at the mid point of the spring , due to a force on system , the spring has got an extension $x_0$ at time $t$ , hence now , the second mass (block) , will lack a displacement of $x_0/2$ .

Therefore total displacement of second mass (block) ,

$d^{\prime }=\frac{F{t}^2}{4m}-\frac{x_0}{2}=\frac{1}{2}(\frac{F{t}^2}{2m}-x_0)$