91Ó°ÊÓ

i) The mass of the block A is,$m_A=1.0\mathrm{kg}$

ii) The mass of the block B is, $m_B=2.0\mathrm{kg}$

iii) The angle of inclination is,$\mathrm{θ}=30°$

iv) The block B has fallen,$d=25\mathrm{cm}=0.25\mathrm{m}$

First, we have to find an increase in the height of block A due to block B. By using the change in gravitational potential and applying conservation of mechanical energy, we can find total kinetic energy.

Formula:

i) The change in gravitational potential is,$\mathrm{Δ}U=-m_{B}gd+m_{A}gh$

ii) The conservation of mechanical energy is,$\mathrm{Δ}{E}_{th}=\mathrm{Δ}K+\mathrm{Δ}U$

Here, if block B falls vertically, then block A must increase its height by

$$\begin{gathered} h=d\sin \mathrm{θ} \\ ⇒h=0.25×\sin 30 \\ ⇒h=0.125m \end{gathered}$$

Therefore, the change in gravitational potential energy is given by,

$\mathrm{Δ}U=-m_{B}gd+m_{A}gh$

Applying conservation of mechanical energy,

$\mathrm{Δ}{E}_{th}=\mathrm{Δ}K+\mathrm{Δ}U=0$

Hence, the change in kinetic energy is given by,

$\mathrm{Δ}K=-\mathrm{Δ}U$

Since, the initial kinetic energy is zero then final kinetic energy is,

$K_f=\mathrm{Δ}K=-\mathrm{Δ}U$

$$\begin{gathered} ⇒K_f=m_{B}gd-m_{A}gh \\ ⇒K_f=\left(2.0×9.8×0.25\right)-\left(1.0×9.8×0.125\right) \\ ⇒K_f=3.7\mathrm{J} \end{gathered}$$