Question

Two blocks with masses $m_{1}=3 \,kg$ and $m_{2}=5\, kg$ are connected by a light string that slides over a frictionless pulley as shown in figure. Initially, $m_{2}$, is held $5\, m$ off the floor while $m_{1}$ is on the floor. The system is then released. At what speed (in $m / s$ ) does $m_{2}$ hit the floor?

Answer 1

The initial and final configurations are shown in the figure. It is convenient to set $U=0$ at the floor. Initially, only $m_{2}$ has potential energy. As it falls, it loses potential energy and gains. kinetic energy. At the same time, $m_{1}$ gains potential energy and kinetic energy. Just before $m_{2}$ lands, it has only kinetic energy. Let $v$ the final speed of each mass. Then, using the law of conservation of mechanical energy.
Final (Just before $m_{2}$ strikes the floor)
$K_{f}+U_{f}=K_{i}+U_{i} $
$\frac{1}{2}\left(m_{1}+m_{2}\right) v^{2}+m_{1} g h=0+m_{2} g h$

$v^{2}=\frac{2\left(m_{2}-m_{1}\right) g h}{m_{1}+m_{2}}$
Putting $m_{1}=3 \,kg ; m_{2}=5 \, kg $;
$ h=5 \,m$ and $g=10 \,m / s ^{2}$
we get $v^{2}=\frac{2(5-3)(10)(5)}{5+3}$
or $v=5 \,m / s$