A chain of metal links with total mass M = 7 kg is coiled up in a tight ball on a low-friction table (Figure 9.52). You pull on a link at one end of the chain with a constant force F= 50 N. Eventually the chain straightens out to its full length = 2.6 m. and you keep pulling until you have pulled your end of the chain a total distance d=4.5 m.

(a) Consider the point particle system. What is the speed of the chain at this instant? (b) Consider the extended system. What is the change in energy of the chain? (c) In straightening out, the links of the chain bang against each other, and their temperature rises. Assume that the process is so fast that there is insufficient time for significant transfer of energy from the chain to the table due to the temperature difference, and ignore the small amountof energy radiated away as sound produced in the collisions among the links. Calculate the increase in internal energy of the chain.

Given data can be listed below,

• Mass,$\mathrm{M}=7\mathrm{kg}$
• Force,$\mathrm{F}=50\hspace{0.33em}\mathrm{N}$
• The chain straightens out to full length,$L=2.6\hspace{0.33em}m$
• Total distance to pull the end of the chain,$\mathrm{d}=4.5\hspace{0.33em}\mathrm{m}$

The translation work done on the chain is given as,

${\mathrm{W}}_{\mathrm{trans}}=\mathrm{F}\left(\mathrm{d}-\frac{\mathrm{L}}{2}\right)$

Substituting 50 N for F, 4.5 m for d, and 2.6 m for L in the above equation

$$\begin{gathered} {\mathrm{W}}_{\mathrm{trans}}=\left(50\mathrm{N}\right)\left(4.5\mathrm{m}-\frac{2.6\mathrm{m}}{2}\right) \\ =160\mathrm{J} \end{gathered}$$

Part a)

By using conservation of energy, the translational work done is equal to translational kinetic energy, which is written as,

$$\begin{gathered} {\mathrm{W}}_{\mathrm{trans}}={\mathrm{KE}}_{\mathrm{trans}} \\ =\frac{1}{2}{\mathrm{Mv}}^2 \end{gathered}$$

Where v is the speed of the chain relative to the center of the chain.

Substituting 160 J for$W_{trans}$ , and 7 kg for M in the above equation, we get

$$\begin{gathered} \left(160\mathrm{J}\right)=\frac{1}{2}\left(7\mathrm{kg}\right){\mathrm{V}}^2 \\ {\mathrm{v}}^2=\frac{2\left(160\mathrm{J}\right)}{\left(7\mathrm{kg}\right)} \\ \mathrm{v}=6.76\mathrm{m}/\mathrm{s} \end{gathered}$$

Thus, the speed of the chain at the instant is 6.76 m/s.

Part b)

The change in energy is equal to the total work done by the chain is given as,

$$\begin{gathered} ∆E=W_{total} \\ =Fd \end{gathered}$$

Substituting 50 N for F, and 4.5 m for d in the above equation

$$\begin{gathered} ∆\mathrm{E}=\left(50\mathrm{N}\right)\left(4.5\mathrm{m}\right) \\ =225\mathrm{J} \end{gathered}$$

Thus, the change in energy of the chain is 225 J.

Part c)

By using the 1^st law of thermodynamics,

$\Delta E=W_{trans}+\Delta Q$

Where is the change in thermal energy.

Substituting 160 J for$W_{trans}$ , and 225 J for in the above equation

$$\begin{gathered} \mathrm{\Delta Q}=\mathrm{\Delta E}-{\mathrm{W}}_{\mathrm{trans}} \\ =225\mathrm{J}-160\mathrm{J} \\ =65\mathrm{J} \end{gathered}$$

Thus, the increase in the thermal energy of the chain is 65 J.