The arrangement shown in Fig. 8-24 is similar to that in Question 6. Here you pull downward on the rope that is attached to the cylinder, which fits tightly on the rod. Also, as the cylinder descends, it pulls on a block via a second rope, and the block slides over a lab table. Again consider the cylinder–rod–Earth system, similar to that shown in Fig. 8-23b. Your work on the system is $\mathbf{200}\mathbf{}\mathbf{J}$.The system does work of $\mathbf{60}\mathbf{}\mathbf{J}$on the block. Within the system, the kinetic energy increases by $\mathbf{130}\mathbf{}\mathbf{J}$and the gravitational potential energy decreases by 20 J. (a) Draw an “energy statement” for the system, as in Fig. 8-23c. (b) What is the change in the thermal energy within the system?

• Total work, $W=200\mathrm{J}$.
• Work on the block$W_b=60\mathrm{J}$.

• Decrease in potential energy, $∆PE=-20\mathrm{J}$.
• Increase in the kinetic energy, role="math" localid="1657185565146" $∆KE=130\mathrm{J}$.

Here the relation between work done and energy can be used to find change in thermal energy. It is known that the energy can neither be created nor be destroyed but it gets transformed into some other form. In this case that is termed as work done. Here the concept of work done on the system due to all energies - potential energy, kinetic energy, and thermal energy can be used.

Formula:

The work done is given by,

$W=∆KE+∆PE+∆E_{th}$

The different types of energies are given, therefore, the energy statement for the system is,

Total work is,$W=200\mathrm{J}$

The change in the potential energy is,$∆PE=-20\mathrm{J}$

The change in the kinetic energy is,$∆KE=130\mathrm{J}$

The change in the thermal energy is,$E_{th}=?$

For the total work done on the system

$W=W_s+W_b$

Where,

W=Total work done

$W_s=$Work done on the system

$W_b=$Work done on the block

As

$$\begin{gathered} W=+200\mathrm{J},W_b=60\mathrm{J} \\ +200\mathrm{J}=W_{\mathit{s}}+60\mathrm{J} \\ {W}_{\mathrm{s}}=140\mathrm{J} \end{gathered}$$

Now,

role="math" localid="1657186630428" $$\begin{gathered} W_s=∆KE+∆PE+∆E_{th} \\ ∆E_{th}=W_s-(∆KE+∆PE) \\ ∆E_{th}=140\mathrm{J}-(130\mathrm{J}-20\mathrm{J}) \\ ∆E_{th}=30\mathrm{J} \end{gathered}$$

Therefore, the change in thermal energy$∆E_{th}$ within the system is$∆E_{th}=30\mathrm{J}$