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.
- Determine the speed of the block A just before hitting the ground.
- Determine the tension of the cord during the motion.
- 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.