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Chapter 6: Problem 24

Calculate the kinetic energy of an object with a mass of \(1.0 \times\) \(10^{-5} \mathrm{~g}\) and a velocity of \(2.0 \times 10^{5} \mathrm{~cm} / \mathrm{s}\)

Short Answer

Expert verified
The kinetic energy of the object with a mass of \(1.0 \times 10^{-8} \mathrm{~kg}\) and a velocity of \(2.0 \times 10^{3} \mathrm{~m/s}\) is \(8.0 \times 10^{-2} \mathrm{~J}\) (Joules).

Step by step solution

01

Convert the given mass and velocity to standard units

To convert mass from grams to kilograms, we will divide by 1000. To convert velocity from centimeters per second to meters per second, we will divide by 100. Mass in kg: \(1.0 \times 10^{-5}\mathrm{~g} = 1.0 \times 10^{-5}\mathrm{~g} \times \frac{1\mathrm{~kg}}{1000\mathrm{~g}} = 1.0 \times 10^{-8}\mathrm{~kg}\) Velocity in m/s: \(2.0 \times 10^{5}\mathrm{~cm/s} = 2.0 \times 10^{5}\mathrm{~cm/s} \times \frac{1\mathrm{~m}}{100\mathrm{~cm}} = 2.0 \times 10^{3}\mathrm{~m/s}\)
02

Calculation of kinetic energy

Now that we have the mass and velocity in standard units, we can use the kinetic energy formula: K.E. = 0.5 * m * v^2 Substitute the values: K.E. = 0.5 * \(1.0 \times 10^{-8}\mathrm{~kg}\) * \((2.0 \times 10^{3}\mathrm{~m/s})^2\)
03

Simplify and compute the kinetic energy

Calculate the square of the velocity and then multiply by the mass and 0.5. K.E. = 0.5 * \(1.0 \times 10^{-8}\mathrm{~kg}\) * \((4.0 \times 10^{6}\mathrm{~(m/s)^2})\) K.E. = \(2.0 \times 10^{-8}\mathrm{~kg} \times 4.0 \times 10^{6}\mathrm{~(m/s)^2}\) K.E. = \(8.0 \times 10^{-2}\mathrm{~J}\) (Joules) The kinetic energy of the object is 8.0 x 10^{-2} Joules.

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Key Concepts

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

Unit Conversion
When working with physical quantities, it's often necessary to convert units to standard forms. This step is crucial as it ensures consistency and accuracy.

In our problem, we start with the mass given in grams and the velocity in centimeters per second. To convert these:
  • Mass: Convert from grams to kilograms. Since 1 kg equals 1000 g, divide the mass by 1000. Our mass of 1.0 x 10^-5 g becomes 1.0 x 10^-8 kg.
  • Velocity: Change from cm/s to m/s. Knowing that 1 m is 100 cm, divide the velocity by 100. So, 2.0 x 10^5 cm/s converts to 2.0 x 10^3 m/s.
This standardization allows us to apply formulas and comprehend results in commonly accepted scientific units, like Joules for kinetic energy.
Kinetic Energy Formula
Kinetic energy is the energy an object possesses because of its motion. Calculating it involves the kinetic energy formula: \(KE = \frac{1}{2} m v^2\). Here:
  • \(m\) represents mass in kilograms (kg).
  • \(v\) is velocity in meters per second (m/s).
Using this, we find the object's kinetic energy by substituting the converted mass and velocity into the formula. This calculation captures how both mass and speed contribute to motion energy. In essence, even small mass gains significant energy at high velocities.
Physics
Physics, the science of matter and energy, relies heavily on formulas like those for kinetic energy to describe and predict natural phenomena. Understanding how objects move and interact involves:
  • Principles of motion, like Newton's laws, impacting how energy manifests.
  • Units such as Joules, which measure energy levels consistently.
  • Conservative systems, meaning energy doesn't vanish but changes form.
In calculating kinetic energy, physics principles help turn data into applicable insights about energy transformations and the forces affecting motion.
Mass and Velocity
Mass and velocity are fundamental quantities in physics, directly influencing kinetic energy. Let's delve into each:
  • Mass: The amount of matter in an object, measured in kilograms. Greater mass means more energy required for movement, thus yielding more kinetic energy for the same velocity.
  • Velocity: This vector quantity combines speed and direction. As velocity squared appears in the kinetic energy formula, an increase in velocity has a much larger effect on energy than an equivalent increase in mass.
In practice, these concepts explain why lighter but faster objects sometimes have comparable or superior energy levels to heavier, slower ones. Understanding these factors reveals how different scenarios affect movement and energy priorities.

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Most popular questions from this chapter

The Ostwald process for the commercial production of nitric acid from ammonia and oxygen involves the following steps: $$ \begin{aligned} 4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) & \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) \\ 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) & \longrightarrow 2 \mathrm{NO}_{2}(g) \\ 3 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) & \longrightarrow 2 \mathrm{HNO}_{3}(a q)+\mathrm{NO}(g) \end{aligned} $$ a. Use the values of \(\Delta H_{\mathrm{f}}^{\circ}\) in Appendix 4 to calculate the value of \(\Delta H^{\circ}\) for each of the preceding reactions. b. Write the overall equation for the production of nitric acid by the Ostwald process by combining the preceding equations. (Water is also a product.) Is the overall reaction exothermic or endothermic?

Consider the following cyclic process carried out in two steps on a gas: Step 1: \(45 \mathrm{~J}\) of heat is added to the gas, and \(10 . \mathrm{J}\) of expansion work is performed. Step 2: \(60 . \mathrm{J}\) of heat is removed from the gas as the gas is compressed back to the initial state. Calculate the work for the gas compression in step \(2 .\)

The standard enthalpy of combustion of ethene gas, \(\mathrm{C}_{2} \mathrm{H}_{4}(g)\), is \(-1411.1 \mathrm{~kJ} / \mathrm{mol}\) at \(298 \mathrm{~K}\). Given the following enthalpies of formation, calculate \(\Delta H_{\mathrm{f}}^{\circ}\) for \(\mathrm{C}_{2} \mathrm{H}_{4}(g)\). $$ \begin{array}{ll} \mathrm{CO}_{2}(g) & -393.5 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{H}_{2} \mathrm{O}(l) & -285.8 \mathrm{~kJ} / \mathrm{mol} \end{array} $$

A balloon filled with \(39.1\) mol helium has a volume of \(876 \mathrm{~L}\) at \(0.0^{\circ} \mathrm{C}\) and \(1.00\) atm pressure. The temperature of the balloon is increased to \(38.0^{\circ} \mathrm{C}\) as it expands to a volume of \(998 \mathrm{~L}\), the pressure remaining constant. Calculate \(q, w\), and \(\Delta E\) for the helium in the balloon. (The molar heat capacity for helium gas is \(20.8 \mathrm{~J} /{ }^{\circ} \mathrm{C} \cdot\) mol. \()\)

Are the following processes exothermic or endothermic? a. When solid \(\mathrm{KBr}\) is dissolved in water, the solution gets coldel b. Natural gas \(\left(\mathrm{CH}_{4}\right)\) is burned in a furnace. c. When concentrated \(\mathrm{H}_{2} \mathrm{SO}_{4}\) is added to water, the solution get very hot. d. Water is boiled in a teakettle.
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