Question

A 1kg block and a 0.5 kg block move on a horizontal frictionless surface. Each block exert a force of 6N on each other. The block move with a uniform acceleration of

$\begin{aligned} & a)3m{{s}^{-2}} \\\ & b)6m{{s}^{-2}} \\\ & c)9m{{s}^{-2}} \\\ & d)12m{{s}^{-2}} \\\ \end{aligned}$

Answer 1

In the above diagram we can see that the both the blocks move together with the same acceleration. It is given to us both the blocks exert a force i.e. normal reaction on each other of 6N. The normal reaction because of 1kg block keeps the block with mass 0.5kg and keeps on moving with the particular acceleration. Hence we will only take into consideration of this case and determine the acceleration of the two blocks.

Formula used:
$F=ma$

Complete step-by-step answer:
It is given in the question that the two blocks apply a force of 6N on each other. This force actually keeps the block with mass 0.5 kg and keeps on moving with the particular acceleration. Let us say a body has mass ‘m’ and acceleration ‘a’, the force (F)experienced by the body is given by,
$F=ma$
The force on the block with mass 0.5kg is given to us as 6N. Therefore its acceleration from the above expression we get as,
$\begin{aligned} & F=ma \\\ & \Rightarrow 6=0.5a \\\ & \Rightarrow a=\dfrac{6}{0.5}=12m{{s}^{-2}} \\\ \end{aligned}$
Therefore both the blocks move with the uniform acceleration of $12m{{s}^{-2}}$. Hence the correct answer of the above question is option d

So, the correct answer is “Option d”.

Note: In the above case it is given to us that both the blocks are in contact. Hence we could conclude that both the blocks move with the same acceleration. It is also to be noted that the normal force is a contact force. It comes into play when both the blocks are in contact and a force is applied on one of the blocks.