Question

Two blocks ($ m = 0.5 kg$ and $ M = 4.5 kg$) are arranged on a horizontal frictionless table as shown in the figure. The coefficient of static friction between the two blocks is $ \frac{3}{7}$. Then the maximum horizontal force that can be applied on the larger block so that the blocks move together is _______ $ N$. (Round off to the nearest integer) [Take $ g$ as $ 9.8 m{s}^{-2}$]

Answer 1

Step 1: Given Data

Mass of the first block $m=0.5kg$

Mass of the second block $M=4.5kg$

Coefficient of friction $\mu =\frac{3}{7}$

Acceleration due to gravity $g=9.8m{s}^{-2}$

Let $N$ be the normal force.

Let the acceleration be $a$

Step 2: Formula Used

Maximum force, $f_{max}=\mu N$

Step 3: Calculate the acceleration

Therefore, the maximum force that can be applied to the first block,

$$\begin{gathered} f_{max}=ma \\ \Rightarrow \mu N=ma \\ \Rightarrow \mu mg=ma \\ \Rightarrow a=\mu g \\ \Rightarrow a=\frac{3}{7}\times 9.8 \\ \Rightarrow a=4.2m{s}^{-2} \end{gathered}$$

Step 4: Calculate the Maximum Force

Therefore, the total maximum force that can be applied to the system,

$$\begin{gathered} F=\left(M+m\right)a \\ =\left(4.5+0.5\right)4.2 \\ =5\times 4.2 \\ =21N \end{gathered}$$

Hence, the maximum horizontal force that can be applied on the larger block so that the blocks move together is $21N$.