Question

In Fig. 8-60, the pulley has negligible mass, and both it and the inclined plane are frictionless. Block A has a mass of 1.0 kg, block B has a mass of 2.0 kg, and angle $\mathit{\theta }$is $\mathbf{30}\mathbf{°}$. If the blocks are released from rest with the connecting cord taut, what is their total kinetic energy when block B has fallen 25 cm?

Short answer

The total kinetic energy is, $K_f=3.7\mathrm{J}$

Step-by-step solution

Given

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,$\theta =30°$

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

Determine the formula for the potential and the mechanical energy:

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,$\Delta U=-m_{B}gd+m_{A}gh$

ii) The conservation of mechanical energy is,$\Delta E_{th}=\Delta K+\Delta U$

Calculate their total kinetic energy when block B has fallen 25 cm 

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

$$\begin{gathered} h=d\sin \theta \\ \Rightarrow h=0.25\times \sin 30 \\ \Rightarrow h=0.125m \end{gathered}$$

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

$\Delta U=-m_{B}gd+m_{A}gh$

Applying conservation of mechanical energy,

$\Delta E_{th}=\Delta K+\Delta U=0$

Hence, the change in kinetic energy is given by,

$\Delta K=-\Delta U$

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

$K_f=\Delta K=-\Delta U$

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