Question:
The heavier block in an Atwood machine has a mass twice that of the lighter one. The tension in the string is $16.0 \mathrm{~N}$ when the system is set into motion. Find the decrease in the gravitational potential energy during the first second after the system is released from rest.
Solution:
Net force is
$\mathrm{T}-2 \mathrm{mg}+2 \mathrm{ma}=0 \ldots \ldots 1$
$\mathrm{T}-\mathrm{mg}-\mathrm{ma}=0 \ldots \ldots 2$
from equation 1 and 2
$a=\frac{T}{4 m}=\frac{16}{4 m}$
$a=\bar{m} \mathrm{~m} / \mathrm{s}^{2}$
Now
$s=u t+\frac{1}{2} a t^{2}$
$s=0+\frac{1}{2} \times \frac{4}{m} \times 1^{2}$
$\mathrm{s}=\frac{2}{m}$
Decrease in P.E. = mgh
P.E. $=(2 \mathrm{~m}-\mathrm{m}) \times 9.8 \times(2 / \mathrm{m})$
P.E. $=19.6 \mathrm{~J}$