Two blocks (m = 0.5, kg and M = 4.5, kg) are | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) For the blocks to move together,the acceleration $a$ of the system must be such that the frictional force on the smaller block $m$ is sufficient t

Vedclass · Accepted Answer

$21$

Vedclass · Answer

(C) For the blocks to move together,the acceleration $a$ of the system must be such that the frictional force on the smaller block $m$ is sufficient to provide its acceleration.The maximum frictional force available is $f_{\max} = \mu N = \mu mg$.The maximum acceleration $a_{\max}$ that the block $m$ can have without slipping is $a_{\max} = \frac{f_{\max}}{m} = \mu g$.Given $\mu = \frac{3}{7}$ and $g = 9.8\, m/s^2$,we have:$a_{\max} = \frac{3}{7} 	imes 9.8 = 3 	imes 1.4 = 4.2\, m/s^2$.Now,for the entire system of mass $(M + m)$,the force $F$ is applied:$F = (M + m) a_{\max}$.Substituting the values:$F = (4.5 + 0.5) 	imes 4.2 = 5 	imes 4.2 = 21\, N$.Thus,the maximum horizontal force is $21\, N$.

Similar Questions

The coefficient of friction between $4 \, kg$ and $5 \, kg$ blocks is $0.2$ and between $5 \, kg$ block and ground is $0.1$ respectively. Choose the correct statements.

$A$ block of mass $m$ is placed on the top of another block of mass $M$ as shown in the figure. The coefficient of friction between them is $\mu$. What is the maximum acceleration with which the block $M$ may move so that $m$ also moves along with it?

The frictional force acting on the $1 \,kg$ block is .................. $N$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module