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Chapter 6: Problem 4

A truck initially traveling at \(60 \mathrm{~km}\) per hour is brought to a complete stop at a traffic light. Does this change violate the law of conservation of energy? Explain.

Short Answer

Expert verified
No, this change does not violate the law of conservation of energy. The kinetic energy of the moving truck is transformed into thermal energy when the truck is stopped by the use of brakes.

Step by step solution

01

Identify the forms of energy involved

Look first at the situation before the truck came to the stop. The truck was moving, so it had kinetic energy, which is the energy of motion. The kinetic energy of the truck can be calculated by the formula \(\frac{1}{2} m v^2\), where \(m\) is the mass of the truck and \(v\) is its speed.
02

Account for the energy after the truck stopped

After the truck comes to a stop, it is no longer moving, so it no longer has kinetic energy. However, the energy has not disappeared. The kinetic energy was transformed into another form of energy. When the truck slows down and stops, the brakes are applied. The friction between the brakes and the wheels generates heat, transforming the kinetic energy into thermal energy.
03

Conclusion

Understand that the law of conservation of energy has not been violated. Even though the truck has stopped and it no longer has kinetic energy, this energy has not disappeared. Instead, it has been transformed into thermal energy. This transformation of energy from one form to another is what the law of conservation of energy is all about and is evidenced by heat generated in the brakes in the process of stopping the truck.

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Most popular questions from this chapter

Define these terms: system, surroundings, open system, closed system, isolated system, thermal energy, chemical energy, potential energy, kinetic energy, law of conservation of energy.

Consider the following data: $$ \begin{array}{lcc} \text { Metal } & \text { Al } & \text { Cu } \\ \hline \text { Mass (g) } & 10 & 30 \\ \text { Specific heat (J/g* } \left.^{\circ} \mathrm{C}\right) & 0.900 & 0.385 \\\ \text { Temperature }\left({ }^{\circ} \mathrm{C}\right) & 40 & 60 \\ \hline \end{array} $$ When these two metals are placed in contact, which of the following will take place? (a) Heat will flow from Al to Cu because Al has a larger specific heat. (b) Heat will flow from \(\mathrm{Cu}\) to \(\mathrm{Al}\) because \(\mathrm{Cu}\) has a larger mass. (c) Heat will flow from Cu to Al because Cu has a larger heat capacity. (d) Heat will flow from Cu to Al because Cu is at a higher temperature. (e) No heat will flow in either direction.

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