A constant force F acts on two different mass | vedclass

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

Question

(C) Given that a constant force $F$ acts on mass $m_1$,the acceleration is $A_1 = \frac{F}{m_1}$,which implies $m_1 = \frac{F}{A_1}$.Similarly,for mas

Vedclass · Accepted Answer

$\frac{A_1 A_2}{A_1 + A_2}$

Vedclass · Answer

(C) Given that a constant force $F$ acts on mass $m_1$,the acceleration is $A_1 = \frac{F}{m_1}$,which implies $m_1 = \frac{F}{A_1}$.Similarly,for mass $m_2$,the acceleration is $A_2 = \frac{F}{m_2}$,which implies $m_2 = \frac{F}{A_2}$.When the same force $F$ acts on the combined mass $(m_1 + m_2)$,the acceleration $A$ is given by $A = \frac{F}{m_1 + m_2}$.Substituting the values of $m_1$ and $m_2$ into the equation:$A = \frac{F}{\frac{F}{A_1} + \frac{F}{A_2}}$$A = \frac{F}{F(\frac{1}{A_1} + \frac{1}{A_2})}$$A = \frac{1}{\frac{A_1 + A_2}{A_1 A_2}}$$A = \frac{A_1 A_2}{A_1 + A_2}$.

$A$ constant force $F$ acts on two different masses $m_1$ and $m_2$ independently and produces accelerations $A_1$ and $A_2$ respectively. When the same force $F$ acts on their combined mass $(m_1 + m_2)$,the acceleration $A$ produced is:

Similar Questions

$A$ force $\vec{F} = a \hat{i} + b \hat{j} + c \hat{k}$ is acting on a body of mass $m$. The body was initially at rest at the origin. The coordinates of the body after time $t$ will be:

The motion of a particle of mass $m$ is described by $y = ut + \frac{1}{2}gt^2$. Find the force acting on the particle.

$A$ block of mass $1 \ kg$ starts from rest at $x=0$ and moves along the $X$-axis under the action of a force $F=kt$, where $t$ is time and $k=1 \ Ns^{-1}$. The distance the block will travel in $6 \ seconds$ is: (in $m$)

$A$ body of mass $5 \ kg$ starts from the origin with an initial velocity $(30 \hat{i} + 40 \hat{j}) \ m/s$. If a constant force $-(\hat{i} + 5 \hat{j}) \ N$ acts on the body,then the time in which the $y$-component of its velocity becomes zero is (in $s$)

$A$ force on an object of mass $100\, g$ is $(10 \hat{i}+5 \hat{j})\ N$. The position of that object at $t= 2\ s$ is $(a \hat{i}+b \hat{j})\, m$ after starting from rest. The value of $\frac{a}{b}$ will be..........

Vedclass Test Series

Exam Paper Generator

Online Exam Module