Three blocks of masses m1, m2 and m3 kg are placed in contact with each other on a frictionless table. A force F is applied on the heaviest mass m1; the acceleration of m3 will be

Question

Three blocks of masses m1, m2 and m3 kg are placed in contact with each other on a frictionless table. A force F is applied on the heaviest mass m1; the acceleration of m3 will be (A) (F/m1) (B) (F/m1 + m2) (C) (F/m2 + m3) (D) (F/m1 + m2 + m3)

Tardigrade Admin · Accepted Answer

The acceleration of mass m3 =common acceleration of the system =( text force applied / text total mass ) =(F/m1+m2+m3)

Q. Three blocks of masses $m_{1}, m_{2}$ and $m_{3}$ kg are placed in contact with each other on a frictionless table. A force $F$ is applied on the heaviest mass $m_{1}$ ; the acceleration of $m_{3}$ will be

$\frac{F}{m _{1}}$

$\frac{F}{m _{1} + m _{2}}$

$\frac{F}{m _{2} + m _{3}}$

$\frac{F}{m _{1} + m _{2} + m _{3}}$

The acceleration of mass $m_{3}$
$=$ common acceleration of the system
$= \frac{force applied}{total mass}$
$= \frac{F}{m _{1} + m _{2} + m _{3}}$

Q. Three blocks of masses m1​,m2​ and m3​ kg are placed in contact with each other on a frictionless table. A force F is applied on the heaviest mass m1​; the acceleration of m3​ will be

m1​F​

m1​+m2​F​

m2​+m3​F​

m1​+m2​+m3​F​