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Sagot :
Given data:
* The mass of the lead is m = 50 kg.
* The angle of the inclined plane with the horizontal is,
[tex]\theta=10^{\circ}[/tex]* The length of the inclined plane is L = 5 m.
* The speed of the lead at the bottom of the plane is v = 2.2 m/s.
* The initial temperature of the lead is,
[tex]T_i=15^{\circ}\text{ C}[/tex]* The specific heat of lead is,
[tex]c_L=130J/kg^{\circ}C[/tex]Solution:
(a). The kinetic energy of the lead at the bottom of the inclined plane is,
[tex]\begin{gathered} K=\frac{1}{2}mv^2 \\ K=\frac{1}{2}\times50\times2.2^2 \\ K=121\text{ J} \end{gathered}[/tex]The diagrammatic representation of the given case is,
From the diagram, the initial height of the block is,
[tex]\begin{gathered} \sin (\theta)=\frac{h}{L} \\ h=L\sin (\theta) \end{gathered}[/tex]Substituting the known values,
[tex]\begin{gathered} h=5\times\sin (10^{\circ}) \\ h=0.87\text{ m} \end{gathered}[/tex]The potential energy at the top of the inclined plane is,
[tex]U=\text{mgh}[/tex]where g is the acceleration due to gravity,
Substituting the known values,
[tex]\begin{gathered} U=50\times9.8\times0.87 \\ U=426.3\text{ J} \end{gathered}[/tex]The lead block is initially at rest at the top of the inclined plane, thus, the kinetic energy at the top of the inclined plane is zero.
The net energy at the top of the inclined plane is,
[tex]\begin{gathered} E_i=U+K_i \\ E_i=U+0 \\ E_i=426.3\text{ J} \end{gathered}[/tex]The height of the lead block at the bottom of the inclined plane is zero, thus, the potential energy at the bottom of the inclined plane is zero.
The net energy at the bottom of the inclined plane is,
[tex]\begin{gathered} E_f=U_f+K \\ E_f=0+121 \\ E_f=121\text{ J} \end{gathered}[/tex]Thus, the energy loss due to the friction is,
[tex]\begin{gathered} E_f-E_i=121-426.3 \\ E=-305.3\text{ J} \end{gathered}[/tex]Here, the negative sign indicates the loss of energy,
The work done by the frictional force is equal to the amount of energy loss during the motion of the lead block down the inclined plane.
Thus, the work done by the friction force on the lead is 305.3 J.
(b). The energy loss takes place in the form of heat.
The amount of heat produced in terms of the mass, temperature, and specific heat is,
[tex]Q=mc_L(T_f-T_i)[/tex]where Q is the amount of heat produced, T_f is the final temperature and T_i is the initial temperature,
Substituting the known values,
[tex]\begin{gathered} 305.3=50\times130\times(T_f-15) \\ 305.3=6500T_f-97500 \\ 6500T_f=305.3+97500 \\ 6500T_f=97805.3 \end{gathered}[/tex]By simplifying,
[tex]\begin{gathered} T_f=\frac{97805.3}{6500} \\ T_f=15.05^{\circ}\text{C} \end{gathered}[/tex]Thus, the final temperature of the lead block is 15.05 degrees celsius.
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