A pulley fixed to the ceiling carries a strin | vedclass

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

Question

(B) Let the masses be $m_1 = 3m$ and $m_2 = m$. The acceleration of the blocks is given by $a = \frac{(m_1 - m_2)g}{m_1 + m_2} = \frac{(3m - m)g}{3m +

Vedclass · Accepted Answer

$g/4$

Vedclass · Answer

(B) Let the masses be $m_1 = 3m$ and $m_2 = m$. The acceleration of the blocks is given by $a = \frac{(m_1 - m_2)g}{m_1 + m_2} = \frac{(3m - m)g}{3m + m} = \frac{2mg}{4m} = \frac{g}{2}$.The acceleration of the centre of mass is $a_{cm} = \frac{m_1 a_1 + m_2 a_2}{m_1 + m_2}$.Taking the downward direction as positive for the heavier mass $(3m)$,its acceleration is $a_1 = g/2$. The lighter mass $(m)$ moves upward,so its acceleration is $a_2 = -g/2$.Substituting these values: $a_{cm} = \frac{3m(g/2) + m(-g/2)}{3m + m} = \frac{1.5mg - 0.5mg}{4m} = \frac{mg}{4m} = \frac{g}{4}$.

$A$ pulley fixed to the ceiling carries a string with blocks of mass $m$ and $3m$ attached to its ends. The masses of the string and pulley are negligible. When the system is released,what is the acceleration of its centre of mass?

Similar Questions

At what value of $m_1$ will the $8\, kg$ mass be at rest?

$A$ mass $M$ is lowered by the help of a string at a constant acceleration of $\frac{g}{2}$ for a distance $x$. What is the work done by the string?

Three masses of $1\, kg$,$6\, kg$,and $3\, kg$ are connected to each other with threads and are placed on a table as shown in the figure. The acceleration with which the system is moving is $....... \, ms^{-2}$. (Take $g = 10\, ms^{-2}$)

Two blocks of masses $m$ and $M$ are connected by means of a metal wire of cross-sectional area $A$ passing over a frictionless fixed pulley as shown in the figure. The system is then released. If $M = 2m$,then the stress produced in the wire is

Vedclass Test Series

Exam Paper Generator

Online Exam Module