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Chapter 5: Problem 21

(a) What is work? (b) How do we determine the amount of work done, given the force associated with the work?

Short Answer

Expert verified
(a) Work is a scalar quantity representing the energy transferred to an object when it is moved in the direction of an applied force. It is calculated as the product of the force and the distance the object moves, accounting only for the component of the force acting along the direction of displacement. (b) To determine the amount of work done (W), given the force (F) and distance (d), we use the formula: \(W = Fd\cos(\theta)\), where \(\theta\) is the angle between the direction of the force and the direction of displacement. If the force is applied in the same direction as the displacement (\(\theta = 0\)), the formula becomes: \(W = Fd\).

Step by step solution

01

(a) Definition of Work

Work is a measure of the energy transferred to an object when it is moved in the direction of the applied force. In physics, work is a scalar quantity that is calculated as the product of the force applied to an object and the distance it moves as a result of that force, only accounting for the component of the force that is acting along the direction of displacement.
02

(b) Determining the Amount of Work Done

To determine the amount of work done on an object given the force F associated with the work, we need two key variables: \(F\) (the force applied in Newtons, often represented as N) and \(d\) (the distance the object moves in meters, represented as m). The formula used to calculate work (W) is: \[W = Fd\cos(\theta)\] Where \(W\) represents the work done in Joules (J), \(\theta\) represents the angle between the direction of the force and the direction of displacement. If the force is applied in the same direction as the displacement, \(\theta = 0\), then the formula becomes: \[W = Fd\] For example, if a force of 10 N is applied to an object and the object moves 5 meters in the direction of the force (so \(\theta = 0\)), The work done would be: \[W = (10 \, N)(5 \, m) = 50 \, J\] Thus, the amount of work done is 50 Joules.

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Key Concepts

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

Energy transfer
Energy transfer refers to the process of moving energy from one place or form to another. When we talk about work in physics, we're discussing the transfer of energy through the application of force.
It's important to understand that work is done when energy is transferred to an object, causing it to move in the direction of the applied force. This transfer of energy results in movement or change, emphasizing the pivotal role energy transfer plays in the concept of work.
  • When you apply a force to move a book across a table, you transfer energy from your hand to the book.
  • This energy makes the book move, demonstrating the concept of energy transfer.
Energy transfer not only happens in mechanical movements but also in other forms such as heat or light. However, when we calculate work, we focus on the energy transferred in direct relation to force and movement in physics.
Force and displacement
Force and displacement are key elements in determining the amount of work done on an object. Force refers to any interaction that, when unopposed, changes the motion of an object.
It can be seen as a push or pull on the object, measured in Newtons (N). Displacement, on the other hand, is the change in position of an object due to a force applied.
  • Displacement is a vector quantity that has both magnitude and direction.
  • In calculating work, only the component of force in the direction of displacement is considered.
The relationship between force and displacement is crucial because work is only done when an object is displaced due to the applied force.
For example, if you push a car with a force of 200 N and it moves 3 meters, the displacement is the distance moved in the direction of the push.
Scalar quantity
In physics, quantities are often classified as either vectors or scalars. A scalar quantity has only magnitude and no direction, unlike a vector which has both magnitude and direction.
Work is a scalar quantity. This means that when calculating work, the direction of the force and displacement doesn't affect the result; only their magnitudes matter.
  • Work is calculated using the formula \(W = Fd\cos(\theta)\), which results in a scalar value measured in Joules.
  • This scalar nature of work makes it independent of the path taken; it depends solely on the initial and final positions.
Understanding that work is a scalar quantity is essential in solving physics problems and understanding energy transactions.

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Most popular questions from this chapter

For each of the following compounds, write a balanced thermochemical equation depicting the formation of one mole of the compound from its elements in their standard states and use Appendix \(\mathrm{C}\) to obtain the value of \(\Delta H_{f}^{\circ}\) : (a) \(\mathrm{NO}_{2}(g)\) (b) \(\mathrm{SO}_{3}(g),(\mathrm{c}) \mathrm{NaBr}(s)\) (d) \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(s)\)

(a) What is the specific heat of liquid water? (b) What is the molar heat capacity of liquid water? (c) What is the heat capacity of \(185 \mathrm{~g}\) of liquid water? (d) How many \(\mathrm{kJ}\) of heat are needed to raise the temperature of \(10.00 \mathrm{~kg}\) of liquid water from \(24.6^{\circ} \mathrm{C}\) to \(46.2^{\circ} \mathrm{C} ?\)

Identify the force present and explain whether work is done when (a) a positively charged particle moves in a circle at a fixed distance from a negatively charged particle; (b) an iron nail is pulled off a magnet.

Ozone, \(\mathrm{O}_{3}(g)\), is a form of elemental oxygen that is important in the absorption of ultraviolet radiation in the stratosphere. It decomposes to \(\mathrm{O}_{2}(g)\) at room temperature and pressure according to the following reaction: $$ 2 \mathrm{O}_{3}(g) \longrightarrow 3 \mathrm{O}_{2}(g) \quad \Delta H=-284.6 \mathrm{~kJ} $$ (a) What is the enthalpy change for this reaction per mole of \(\mathrm{O}_{3}(g) ?(\mathbf{b})\) Which has the higher enthalpy under these condi- $$ \text { tions, } 2 \mathrm{O}_{3}(g) \text { or } 3 \mathrm{O}_{2}(g) ? $$

Three common hydrocarbons that contain four carbons are listed here, along with their standard enthalpies of formation: (a) For each of these substances, calculate the molar enthalpy of combustion to \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(l) .\) (b) Calculate the fuel value in \(\mathrm{kJ} / \mathrm{g}\) for each of these compounds. \((\mathrm{c})\) For each hydrocarbon, determine the percentage of hydrogen by mass. (d) By comparing your answers for parts (b) and (c), propose a relationship between hydrogen content and fuel value in hydrocarbons.
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