Question

The heavier block is an Atwood machine that 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?

Answer 1

Step 1: Given data:

Given that,

Tension $\mathrm{T}=16.0\mathrm{N}$

Step 2: Formula for calculating the decreased potential energy:

The formula for the decreased potential energy can be given as

$\mathrm{P}.\mathrm{E}=\mathrm{mg\Delta h}$

Step 3: Expression for the equation of motion:

The equation of motion is basically the equations used to derive components such as displacement(s), velocity (initial and final), time(t) and acceleration(a), which can be given as:

$\mathrm{s}=\mathrm{ut}+\frac{1}{2}{\mathrm{at}}^2$

Where,$\mathrm{s}=$ displacement, $\mathrm{u}=$ velocity, $\mathrm{t}=$ time, and $\mathrm{a}=$ acceleration

The above equation of motion is called the second equation of motion.

Step 4: Calculating the value of $\mathrm{\Delta h}$:

We know,

$$\begin{gathered} \mathrm{T}-2\mathrm{mg}+2\mathrm{ma}=0....\left(\mathrm{I}\right) \\ \mathrm{T}-\mathrm{mg}-\mathrm{ma}=0....\left(\mathrm{II}\right) \end{gathered}$$

From equation $\left(\mathrm{I}\right)\mathrm{and}\left(\mathrm{II}\right)$

$$\begin{gathered} 3\mathrm{ma}-\mathrm{mg}=0 \\ \mathrm{g}=3\mathrm{a} \\ \mathrm{a}=\frac{\mathrm{g}}{3} \end{gathered}$$

Now, put the value of g in the equation $\left(\mathrm{II}\right)$

$$\begin{gathered} \mathrm{T}-3\mathrm{ma}-\mathrm{ma}=0 \\ \mathrm{T}=4\mathrm{ma} \\ \mathrm{a}=\frac{\mathrm{T}}{4\mathrm{m}} \end{gathered}$$

Now, from the equation of motion

At $\mathrm{t}=1,$

$$\begin{gathered} \mathrm{s}=\mathrm{ut}+\frac{1}{2}{\mathrm{at}}^2 \\ \mathrm{s}=0+\frac{1}{2}\times \frac{\mathrm{T}}{4\mathrm{m}}\times {\left(1\right)}^2 \\ \mathrm{s}=\frac{16}{8\mathrm{m}} \\ \mathrm{s}=\frac{2}{\mathrm{m}} \end{gathered}$$

Thus, a change in the height of the block will be;

$\mathrm{\Delta h}=\mathrm{s}$

Step 4: calculating the value of the decreased potential energy :

Now, decrease potential energy can be calculated as:

$$\begin{gathered} \mathrm{P}.\mathrm{E}=\mathrm{mg\Delta h} \\ \mathrm{P}.\mathrm{E}=\mathrm{m}\times 9.8\times \frac{2}{\mathrm{m}} \\ \mathrm{P}.\mathrm{E}=19.6\mathrm{J} \end{gathered}$$

Hence, the decreased gravitational potential energy is $19.6\mathrm{J}$.