Question

A $4kg$ block is on a smooth horizontal table. The block is connected to a second block of mass $1kg$ by a massless flexible taut cord that passed over a frictionless pulley. The $1kg$ block is $1m$ above the floor. The two blocks are released from rest. With what speed does that $1kg$ block hit the ground?

Answer 1

Force on $1kg$ block, $F=mg$ (vertical)

This force give acceleration to the system

$F=(M+m)a$

$a=\frac{F}{M+m}=\frac{mg}{M+m}=\frac{1\times 10}{4+1}=2.5m{s}^{-2}$

Apply kinematic equation of motion,

$v^2-u^2=2as$

$v=\sqrt{2as}=\sqrt{2\times 2.5\times 1}=2.23m{s}^{-1}$

Hence, at velocity $2.23m{s}^{-1}$ it will strike to ground.