Question

Two objects are connected by a light string passing over a light, frictionless pulley as shown in the figure. The object of mass $m_{1}=5.0\, kg$ is released from rest at a height $h=4.0\, m$ above the table. Find the maximum height above the table to which the $3.0- kg$ object rises (in $m$ ).

Answer 1

We assign height $y=0$ to the table top.
Using conservation of energy for the system of the Earth and the two objects:

Then total mechanical energy of the system remains constant and the energy version of the isolated system model gives
$\left(K_{A}+K_{B}+U_{g}\right)_{i}=\left(K_{A}+K_{B}+U_{g}\right)_{f a}$
At the initial point, $K_{A i}$ and $K_{B i}$ are zero and we define the gravitational potential energy of the system as zero. Thus the total initial energy is zero, and we have
$0=\frac{1}{2}\left(m_{1}+m_{2}\right) v_{f a}^{2}+m_{2} g h+m_{1} g(-h)$
Here we have used the fact that because the cord does not stretch, the two blocks have the same speed. The heavier mass moves down, losing gravitational potential energy, as the lighter mass moves up, gaining gravitational potential energy.
Simplifying, $\left(m_{1}-m_{2}\right) g h=\frac{1}{2}\left(m_{1}+m_{2}\right) v_{f a}^{2}$
$\Rightarrow v_{f a}=\sqrt{\frac{2\left(m_{1}-m_{2}\right) g h}{\left(m_{1}+m_{2}\right)}}$
$=\sqrt{\frac{2(5.0 kg -3.0 kg ) 10(4.0 m )}{(5.0 kg +3.0 kg )}}=\sqrt{20} m / s$
Now we apply conservation of mechanical energy for the system of the $3.0- kg$ object and the Earth during the time interval between the instant when the string goes slack and the instant at which the $3.0- kg$ object reaches its highest position in its free fall.
$\Delta K+\Delta U=0$
$\Rightarrow \Delta K=-\Delta U$
$0-\frac{1}{2} m_{2} v^{2}=-m_{2} g \Delta v$
$\Rightarrow \Delta y=\frac{v^{2}}{2 g}=1.0\, m$
Hence, $y_{\max }=4.0 m +\Delta y=5.0\,m$