# Dynamics work and energy problem

Blocks A and B weigh 30lb and 15lb respectively, and they are both at a height 8ft above the ground when the system is released from rest. Assume the pulley is frictionless.

1. Determine the speed of the block A just before hitting the ground.
2. Determine the tension of the cord during the motion.
3. Determine the maximum height reached by block B So far for part 1 I've tried to apply the conservation of energy. Using mgh only for the initial because velocity is 0 and the kinetic energy of both blocks, plus the potential energy of block B at height 16ft for final.

This creates:

$$(30+45)8=\frac{1}{2}(\frac{30}{32.2})v^{2}+\frac{1}{2}\frac{15}{32.2}v^{2}+15(16)$$

Which simplifies to $$v=13.10 ft/s$$

For part 2,

using $$T_{1}+U_{1-2}=T_{2}$$

$$\Rightarrow T_{1}+(W_{A}-F_{A})h=T_{2}$$

$$= F_{A}= W_{A}-\frac{T_{2}}{h}=30-\frac{(.5)(.932)(13.10)^2}{8}=20.003lb$$

I would use the same process for the tension on B.

So to my actual question, is this the correct and easiest way to approach this problem? I don't have access the the solution and am looking for some feedback.

$$F =m\alpha, \quad \text{F is the difference between the 2 masses}, \ (m_1-m_2)g= (m_1+m_2)\alpha \\ \alpha = \frac{(m_1-m_2)g}{(m_1+m_2)}=32(30-15)/(30+15)=32\cdot15/45=32/3\text {ft.s^2}= 10.66$$
$$T_{cord }= m_1g+\alpha =m_2\alpha =30\cdot 10.66=320lbs$$ $$v_{final}=\sqrt{2\alpha H}= 13.05 ft.s$$
Maximum height reached by block B is $$16+ 13.05^2/ 2g=16+13.05^2/2*32=18.66ft$$