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If the mass of a truck is doubled - for example, when it is loaded - by what factor does the kinetic energy of the truck increase? By what factor does the kinetic energy change if the mass is one-tenth of the original mass?

Short Answer

Expert verified
The kinetic energy of the truck increases by a factor of 2 when the mass is doubled and decreases to one-tenth when the mass is one-tenth the original.

Step by step solution

01

Introduction to Kinetic Energy

Kinetic energy (KE) of an object is given by the formula, KE = \( \frac{1}{2}mv^2 \), where \(m\) is the mass of the object and \(v\) is the velocity of the object. The kinetic energy is directly proportional to the mass of the object, meaning if the mass increases, the kinetic energy increases, provided the velocity remains constant.
02

Calculate the Kinetic Energy Change for Doubled Mass

If the mass of the truck is doubled, we represent the new mass as \(2m\). Substituting in the kinetic energy formula, we get the new kinetic energy as \(KE_{new} = \frac{1}{2}(2m)v^2 = 2(\frac{1}{2}mv^2)\). This means that when the mass is doubled, the kinetic energy is also doubled.
03

Calculate the Kinetic Energy Change for One-Tenth Mass

Similarly, if the mass of the truck is one-tenth of the original mass, the new mass is represented as \(\frac{1}{10}m\). The new kinetic energy is given by \(KE_{new} = \frac{1}{2}(\frac{1}{10}m)v^2 = \frac{1}{10}(\frac{1}{2}mv^2)\), which means the kinetic energy is one-tenth of the original kinetic energy when the mass is reduced to one-tenth.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Kinetic Energy Formula
Kinetic energy is a form of energy associated with the motion of an object. The formula to calculate kinetic energy (KE) is given by \( KE = \frac{1}{2}mv^2 \), where \( m \) is the mass of the object in kilograms and \( v \) is its velocity in meters per second. This equation reveals the direct relationship between an object's kinetic energy and its mass and velocity. Understanding this fundamental formula is essential for solving a wide range of physics problems related to motion.

For instance, when you throw a ball, the ball's kinetic energy is determined by how heavy the ball is and how fast it's moving. As per the kinetic energy formula, even a slight increase in velocity will result in a significant increase in kinetic energy, due to the velocity being squared in the formula. This concept demonstrates why high-speed objects like bullets or racing cars possess such large amounts of kinetic energy.
Mass-Velocity Relationship
In the context of kinetic energy, there is a pivotal mass-velocity relationship to comprehend. Kinetic energy is directly proportional to the mass of an object when its velocity remains constant. This implies that if you adjust the mass of an object, keeping the velocity unchanged, the kinetic energy will change in direct proportion to the mass.

Practical Implications

If a 5 kg object moving at a constant velocity of 10 m/s has a certain kinetic energy, increasing the mass to 10 kg, while keeping the velocity the same, will double the kinetic energy. On the other hand, reducing the mass to 0.5 kg will decrease the kinetic energy by 90%, because the kinetic energy is scaled by the same factor as the change in mass. This mass-velocity relationship is a key concept for understanding how different factors influence the motion and energy of objects.
Kinetic Energy Proportionality
Kinetic energy's proportionality can be seen in how changes to an object's mass or velocity affect its kinetic energy. Specifically, kinetic energy is directly proportional to the square of the velocity and directly proportional to the mass of the object. This means that kinetic energy will quadruple if the velocity is doubled (because \( 2^2 = 4 \) ), and will decrease to one-fourth if the velocity is halved (because \( \frac{1}{2}^2 = \frac{1}{4} \)).

Velocity's Powerful Impact

Given that kinetic energy is proportional to the square of velocity, even small changes in speed can have substantial effects on kinetic energy. This is especially important in safety considerations for vehicles, where the energy that needs to be managed in the event of a collision is much greater at higher speeds. Understanding this relationship helps explain concepts like stopping distance and impact forces in car crashes, emphasizing the importance of controlled speed to ensure safety.

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Most popular questions from this chapter

Chlorofluoromethanes (CFMs) are carbon compounds of chlorine and fluorine and are also known as Freons. Examples are Freon-11 \(\left(\mathrm{CFCl}_{3}\right)\) and Freon-12 \(\left(\mathrm{CF}_{2} \mathrm{Cl}_{2}\right),\) which were used as aerosol propellants. Freons have also been used in refrigeration and air-conditioning systems. In 1995 Mario Molina, F. Sherwood Rowland, and Paul Crutzen were awarded the Nobel Prize mainly for demonstrating how these and other CFMs contribute to the "ozone hole" that develops at the end of the Antarctic winter. In other parts of the world, reactions such as those shown below occur in the upper atmosphere where ozone protects the earth's inhabitants from harmful ultraviolet radiation. In the stratosphere CFMs absorb high-energy radiation from the sun and split off chlorine atoms that hasten the decomposition of ozone, \(\mathrm{O}_{3}\). Possible reactions are $$ \begin{array}{ll} \mathrm{O}_{3}(g)+\mathrm{Cl}(g) \longrightarrow \mathrm{O}_{2}(g)+\mathrm{ClO}(g) & \Delta H^{\circ}=-126 \mathrm{~kJ} \\ \mathrm{ClO}(g)+\mathrm{O}(g) \longrightarrow \mathrm{Cl}(g)+\mathrm{O}_{2}(g) & \Delta H^{\circ}=-268 \mathrm{~kJ} \\ \mathrm{O}_{3}(g)+\mathrm{O}(g) \longrightarrow 2 \mathrm{O}_{2}(g) & \Delta H^{\circ}=? \end{array} $$ The \(\mathrm{O}\) atoms in the second equation come from the breaking apart of \(\mathrm{O}_{2}\) molecules caused by ultraviolet radiation from the sun. Use the first two equations to calculate the value of \(\Delta H^{\circ}\) (in kilojoules) for the last equation, the net reaction for the removal of \(\mathrm{O}_{3}\) from the atmosphere.

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