Question

In Fig. 8-23a, you pull upward on a rope that is attached to a cylinder on a vertical rod. Because the cylinder fits tightly on the rod, the cylinder slides along the rod with considerable friction. Your force does work $\mathbf{}\mathbf{W}\mathbf{=}\mathbf{+}\mathbf{100}\mathbf{}\mathbf{J}$on the cylinder–rod–Earth system (Fig. 8-23b).An “energy statement” for the system is shown in Fig. 8-23c: the kinetic energy K increases by 50J, and the gravitational potential energy ${\mathbf{U}}_{\mathbf{g}}$increases by 20 J. The only other change in energy within the system is for the thermal energy${\mathbf{E}}_{\mathbf{th}}$.What is the change ${\mathbf{\Delta E}}_{\mathbf{th}}$?

Short answer

The change in ${\mathrm{\Delta E}}_{\mathrm{th}}$is$30\mathrm{J}$

Step-by-step solution

Given information

• Work on the cylinder-rod-earth system is$\mathrm{W}=+100\mathrm{J}$ .

• The gravitational potential energy increase is,$\mathrm{\Delta PE}=20\hspace{0.33em}\mathrm{J}$.
• The kinetic energy increase is, $\mathrm{\Delta KE}=50\hspace{0.33em}\mathrm{J}$.

To understand the concept

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,

$\mathrm{W}=\mathrm{\Delta KE}+\mathrm{\Delta PE}+{\mathrm{\Delta E}}_{\mathrm{th}}$

To find the change in ∆Eth

We have

$\mathrm{W}=\mathrm{\Delta KE}+\mathrm{\Delta PE}+{\mathrm{\Delta E}}_{\mathrm{th}}$

As given the work, kinetic and potential energy put in the above equation. we get,

$$\begin{gathered} 100\hspace{0.33em}\mathrm{J}=50\hspace{0.33em}\mathrm{J}+20\hspace{0.33em}\mathrm{J}+{\mathrm{\Delta E}}_{\mathrm{th}} \\ {\mathrm{\Delta E}}_{\mathrm{th}}=+100\hspace{0.33em}\mathrm{J}-50\hspace{0.33em}\mathrm{J}-20\hspace{0.33em}\mathrm{J} \\ {\mathrm{\Delta E}}_{\mathrm{th}}=+30\hspace{0.33em}\mathrm{J} \end{gathered}$$

Therefore, the change${\mathrm{\Delta E}}_{\mathrm{th}}$ is$30\hspace{0.33em}\mathrm{J}$ .