A block of mass m is connected to another block of mass M by a massless spring of spring constant k. The blocks are kept on a smooth horizontal plane. Initially, the blocks are at rest and the spring is unstretched. Then a constant force F starts acting on the block of mass M to pull it. The force acting on the block of mass m is

Question

A block of mass m is connected to another block of mass M by a massless spring of spring constant k. The blocks are kept on a smooth horizontal plane. Initially, the blocks are at rest and the spring is unstretched. Then a constant force F starts acting on the block of mass M to pull it. The force acting on the block of mass m is (A) (M F/m+M) (B) (m F/M) (C) ((M+m) F/m) (D) (m F/m+M)

Tardigrade Admin · Accepted Answer

Acceleration of the system of two masses is a=(F/m+M) Force on the block of mass m=m a=(m F/m+M)

$\frac{MF}{m + M}$

$\frac{m F}{M}$

$\frac{( M + m ) F}{m}$

$\frac{m F}{m + M}$

Acceleration of the system of two masses is
$a = \frac{F}{m + M}$
Force on the block of mass
$m = ma = \frac{m F}{m + M}$

m+MMF​

MmF​

m(M+m)F​

m+MmF​

Acceleration of the system of two masses is a=m+MF​ Force on the block of mass m=ma=m+MmF​

$\frac{MF}{m + M}$

$\frac{m F}{M}$

$\frac{( M + m ) F}{m}$

$\frac{m F}{m + M}$

Acceleration of the system of two masses is
$a = \frac{F}{m + M}$
Force on the block of mass
$m = ma = \frac{m F}{m + M}$