Two bodies of mass 6 kg and 4 kg are tied t | vedclass

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

Question

(C) Let the acceleration of the system be $a$ and the tension in the string be $T$.For the $6 \ kg$ mass hanging vertically,the equation of motion is:

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(C) Let the acceleration of the system be $a$ and the tension in the string be $T$.For the $6 \ kg$ mass hanging vertically,the equation of motion is: $6g - T = 6a \quad \dots(i)$For the $4 \ kg$ mass on the frictionless table,the equation of motion is: $T = 4a \quad \dots(ii)$Adding equations $(i)$ and $(ii)$,we get:$6g - T + T = 6a + 4a$$6g = 10a$Substituting $g = 10 \ ms^{-2}$:$6 	imes 10 = 10a$$60 = 10a$$a = 6 \ ms^{-2}$Thus,the acceleration of the $6 \ kg$ mass is $6 \ ms^{-2}$.

Two bodies of mass $6 \ kg$ and $4 \ kg$ are tied to a string as shown in the adjoining figure. If the table and pulley are frictionless,then the acceleration of the $6 \ kg$ mass will be $........ \ ms^{-2}$ $\left(g=10 \ ms^{-2}\right)$.

Similar Questions

In the adjoining figure,the tension in the string connecting blocks $A$ and $B$ is:

In the given figure,two masses $m_1$ and $m_2$ $(m_2 > m_1)$ are at rest in an equilibrium position. Find the tension in string $AB$.

For the given system,the acceleration of the system is ........... $m/s^2$.

Two masses $M_1$ and $M_2$ are accelerated uniformly on a frictionless surface as shown in the figure. The ratio of the tensions $T_1/T_2$ is:

Two masses $m_1 = 5\,kg$ and $m_2 = 4.8\,kg$ tied to a string are hanging over a light frictionless pulley. What is the acceleration of the masses when they are free to move (in $,m/s^2$)? $(g = 9.8\,m/s^2)$

Vedclass Test Series

Exam Paper Generator

Online Exam Module