A constant force F is applied in the horizont | vedclass

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

Question

(B) The system consists of three blocks of masses $M$,$m$,and $M'$ moving together with a common acceleration $a$ on a smooth surface.The total mass o

Vedclass · Accepted Answer

$N > N'$

Vedclass · Answer

(B) The system consists of three blocks of masses $M$,$m$,and $M'$ moving together with a common acceleration $a$ on a smooth surface.The total mass of the system is $M_{total} = M + m + M'$.The acceleration of the system is $a = \frac{F}{M + m + M'}$.To find the contact force $N$ between $M$ and $m$,we consider the free body diagram of the blocks $m$ and $M'$ together. The force $N$ pushes them forward:$N = (m + M') a = \frac{(m + M') F}{M + m + M'}$.To find the contact force $N'$ between $m$ and $M'$,we consider the free body diagram of block $M'$. The force $N'$ pushes it forward:$N' = M' a = \frac{M' F}{M + m + M'}$.Comparing $N$ and $N'$:Since $m > 0$,it follows that $(m + M') > M'$.Therefore,$N > N'$.

$A$ constant force $F$ is applied in the horizontal direction as shown in the figure. If the contact force between $M$ and $m$ is $N$ and the contact force between $m$ and $M'$ is $N'$,then:

Similar Questions

If $\frac{2 x^3+x^2-5}{x^4-25}=\frac{A x+B}{x^2-5}+\frac{C x+1}{x^2+5}$,then $(A, B, C)$ equals to

$A$ line segment $AM = a$ moves in the $XOY$ plane such that $AM$ is parallel to the $X$-axis. If $A$ moves along the circle $x^2 + y^2 = a^2$,then the locus of $M$ is

$A$ steel rod $100 \, cm$ long is clamped at its middle. The fundamental frequency of longitudinal vibrations of the rod is $2.53 \, kHz$. What is the speed of sound in steel in $km/sec$?

$CsBr_3$ contains

$\lim _{x \rightarrow 0} \frac{4^x-9^x}{x(4^x+9^x)}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module