# Question #f158f

Dec 13, 2015

$f = 25 N$

#### Explanation:

Consider the following image, representing the situation

There is a rather easy method to solve this, but one must make a very simple assumption, that the two blocks move together under this acceleration, ie they have the same acceleration

Under this assumption, the two blocks can be considered one composite( this is only true under the above assumption). Applying newtons second law on the composite system
$F = \left({m}_{1} + {m}_{2}\right) a$, again I cannot reiterate how important this assumption is, without that, I couldnt have written that equation. Where $a$ is the acceleration on the system

Therefore with $F = 40 N$, ${m}_{1} = 5$Kg, ${m}_{2} = 3$Kg, we get
$a = 5 \frac{m}{s} ^ 2$

But the only force on the $5$Kg block is due to the frictional force from the $3$Kg, evident from the free body diagram of the 5Kg block. Therefore for the 5Kg block
frictional force=$f = {m}_{1} a$=$5 \setminus \times 5$=$25 N$

Note: one could have as well solved the problem by drawing free body diagrams for both the blocks, but that would be rather messy, taking advantage of one assumption, makes the problem much easier.