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Due to application of force tow blocks of mass 1Kg and 0.5Kg move together. Each block exerts a force of 6N on each other. What is the acceleration by which both the blocks move?

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  • $\begingroup$ I think it depends on the direction of movement and position of blocks. Can You add more info or that's all You got? $\endgroup$
    – Wojciech
    Feb 28, 2014 at 9:01
  • $\begingroup$ What did you try? $\endgroup$
    – Bernhard
    Feb 28, 2014 at 9:56

2 Answers 2

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There is two situations here:

1)The $m_{s}=0.5kg$ mass is in front of the $m_{b}=1 kg$ mass (applied force is applied to 1kg block directly).

2)The $m_{b}=1 kg$ mass is in front of the $m_{s}=0.5 kg$ mass (applied force is applied to 0.5 kg block directly).

Case 1:

Since both blocks apply 6N of force on each other. We know that 6N o force is applied on the front block. Using Newtons 2nd law we get:

$\begin{align} \sum F=& m_{s}a \\\ 6N=&(0.5kg )a \\\ a=&6N/0.5kg \\\ a=&12m/s^2 \end{align}$

Going from here we can find the total applied force as well using the same thing just with total mass as I will call $M=m_{s}+m_{b}$

$\begin{align} \sum F=& Ma \\\ F=&(1.5 kg )12m/s^2 \\\ F=18N \end{align}$

Case 2:

Same thing as case 1 but blocks are switched.

$\begin{align} \sum F=& m_{b}a \\\ 6N=&(1 kg )a \\\ a=&6N/1kg \\\ a=&6m/s^2 \end{align}$

Going from here we can find the total applied force (again) as well using the same thing just with total mass as I will call $M=m_{s}+m_{b}$

$\begin{align} \sum F=& Ma \\\ F=&(1.5 kg )12m/s^2 \\\ F=18N \end{align}$

You should always draw a free body diagram label all forces and apply Newtons laws. Also, I think this is a homework question and maybe voted down (not by me). Hope this helps!

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Visualise the scene, suppose the force $F$ is acting on the bigger block($1kg$) towards right. The smaller block($0.5 kg$) is to the right of the bigger block. Both are moving towards right with an acceleration $a$(suppose).

Consider the smaller block, The only horizontal force acting on it is

$Normal force = 6N$

Due to this force it is moving towards right with the acceleration $a$.

Now,

$Normal force = 0.5 kg * a$

$=> 6N = 0.5 kg * a$

$=> a = 12 m/s$

This is the acceleration of both the blocks. You can also figure out the force $F$.

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