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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}$]

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Solution

Step 1: Given Data

Mass of the first block m=0.5kg

Mass of the second block M=4.5kg

Coefficient of friction μ=37

Acceleration due to gravity g=9.8ms-2

Let N be the normal force.

Let the acceleration be a

Step 2: Formula Used

Maximum force, fmax=μN

Step 3: Calculate the acceleration

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

fmax=maμN=maμmg=maa=μga=37×9.8a=4.2ms-2

Step 4: Calculate the Maximum Force

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

F=M+ma=4.5+0.54.2=5×4.2=21N

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


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