According to the description given in the photo, the attached figure represents the problem graphically for the Atwood machine.
To solve this problem we must apply the concept related to the conservation of energy theorem.
PART A ) For energy conservation the initial kinetic and potential energy will be the same as the final kinetic and potential energy, so



PART B) Replacing the values given as,




Therefore the speed of the masses would be 1.8486m/s
Answer:
2. at the lowest point
Explanation:
The motion of the pendulum is a continuous conversion between kinetic energy (KE) and gravitational potential energy (GPE). This is because the mechanical energy of the pendulum, which is sum of KE and GPE, is constant:
E = KE + GPE = const.
Therefore, when KE is maximum, GPE is minimum, and viceversa.
So, the point of the motion where the KE is maximum is where the GPE is minimum: and since the GPE is directly proportional to the heigth of the bob:

we see that GPE is minimum when the bob is at the lowest point,so the correct answer is
2. at the lowest point
<em>Energy</em><em> </em><em>can</em><em> </em><em>neither </em><em>be</em><em> </em><em>created </em><em>nor</em><em> </em><em>be</em><em> </em><em>destroyed</em><em> </em><em>but</em><em> </em><em>can</em><em> </em><em>be</em><em> </em><em>converted</em><em> </em><em>from</em><em> </em><em>one</em><em> </em><em>form</em><em> </em><em>to</em><em> </em><em>another </em><em>.</em>
You can make sure there's no change in volume by keeping
your gas in a sealed jar with no leaks. Then you can play with
the temperature and the pressure all you want, and you'll know
that the volume is constant.
For 'ideal' gases,
(pressure) times (volume) is proportional to (temperature).
And if volume is constant, then
(pressure) is proportional to (temperature) .
So if you increase the temperature from 110K to 235K,
the pressure increases to (235/110) of where it started.
(400 kPa) x (235/110) = 854.55 kPa. (rounded)
Obviously, choice-b is the right one, but
I don't know where the .46 came from.