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Chapter 2: Problem 19

An object of mass \(10 \mathrm{~kg}\), initially at rest, experiences a constant horizontal acceleration of \(4 \mathrm{~m} / \mathrm{s}^{2}\) due to the action of a resultant force applied for \(20 \mathrm{~s}\). Determine the total amount of energy transfer by work, in kJ.

Short Answer

Expert verified
32 kJ

Step by step solution

01

Calculate the Force

Use Newton's second law, which states that force equals mass times acceleration: \[ F = ma \] Given mass \( m = 10 \text{ kg} \) and acceleration \( a = 4 \text{ m/s}^2 \), \[ F = 10 \text{ kg} \times 4 \text{ m/s}^2 = 40 \text{ N} \]
02

Calculate the Distance Traveled

Use the equation of motion to find the distance traveled under constant acceleration: \[ d = \frac{1}{2} a t^2 \] With initial speed \( u = 0 \text{ m/s} \), acceleration \( a = 4 \text{ m/s}^2 \), and time \( t = 20 \text{ s} \): \[ d = \frac{1}{2} \times 4 \text{ m/s}^2 \times (20 \text{ s})^2 \] \[ d = 2 \times 400 \text{ m} = 800 \text{ m} \]
03

Calculate the Work Done

Work done (which is energy transfer by work) by the force is given by: \[ W = F \times d \] From Step 1, \( F = 40 \text{ N} \) and from Step 2, \( d = 800 \text{ m} \): \[ W = 40 \text{ N} \times 800 \text{ m} = 32,000 \text{ J} \]
04

Convert Joules to Kilojoules

Since we need the answer in kilojoules (kJ), convert joules to kilojoules: \[ 32,000 \text{ J} = 32 \text{ kJ} \]

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Key Concepts

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

Newton's second law
Newton's second law is a fundamental principle that explains how objects accelerate. It states that the force acting on an object is equal to the mass of that object multiplied by its acceleration. In mathematical terms, this is written as: \( F = ma \).

This law is crucial for understanding how forces cause changes in motion. In our exercise, the object has a mass of 10 kg and experiences a constant acceleration of 4 m/s虏. Using Newton's second law, we calculate the force as \( 40 \text{ N} \). Knowing this force is essential for solving the problem because it helps us find the work done later on.
constant acceleration
Constant acceleration means an object's velocity is changing at a uniform rate. This concept is vital to figuring out how far an object travels when it starts from rest or is already moving.

In our exercise, the object undergoes a constant acceleration of 4 m/s虏 due to the applied force. The time this force is applied is 20 seconds. Using the equation of motion, we can calculate the distance traveled by considering this constant acceleration. It's important to note that because the acceleration is constant, the equations we use are simplified and lead to straightforward calculations.

equations of motion
The equations of motion are formulas that relate an object's displacement, velocity, acceleration, and time. For motion with constant acceleration, the second equation of motion is particularly useful: \( d = \frac{1}{2} a t^2 \).

Given the initial speed (u) is 0 m/s, acceleration (a) of 4 m/s虏, and time (t) of 20 seconds, we can calculate the distance (d) the object travels. Substituting the values, \( d = \frac{1}{2} \times 4 \text{ m/s}^2 \times (20 \text{ s})^2 \), gives us 800 meters.

Understanding and applying the correct equation of motion are key steps in determining the distance and thus solving for the total work done.
work-energy principle
The work-energy principle states that the work done by forces on an object results in a change in its energy. The work done by a force can be calculated using: \( W = F \times d \).

In our example, the force (F) calculated using Newton's second law is 40 N, and the distance (d) from the equations of motion is 800 m. Plugging these values into the formula, we get: \( W = 40 \text{ N} \times 800 \text{ m} = 32,000 \text{ J} \), which converts to 32 kJ.

This tells us the amount of energy transferred by the force over the given distance, providing a complete understanding of the energy exchange in this scenario.

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Most popular questions from this chapter

A gas is contained in a vertical piston-cylinder assembly by a piston with a face area of 40 in. \(^{2}\) and weight of \(100 \mathrm{lbf}\). The atmosphere exerts a pressure of \(14.7 \mathrm{lbf} / \mathrm{in}^{2}\) on top of the piston. A paddle wheel transfers 3 Btu of energy to the gas during a process in which the elevation of the piston increases by \(1 \mathrm{ft}\). The piston and cylinder are poor thermal conductors, and friction between them can be neglected. Determine the change in internal energy of the gas, in Btu.

An automobile having a mass of \(900 \mathrm{~kg}\) initially moves along a level highway at \(100 \mathrm{~km} / \mathrm{h}\) relative to the highway. It then climbs a hill whose crest is \(50 \mathrm{~m}\) above the level highway and parks at a rest area located there. For the automobile, determine its changes in kinetic and potential energy, each in kJ. For each quantity, kinetic energy and potential energy, specify your choice of datum and reference value at that datum. Let \(g=9.81 \mathrm{~m} / \mathrm{s}^{2}\).

A gas contained in a piston-cylinder assembly undergoes two processes, \(\mathrm{A}\) and \(\mathrm{B}\), between the same end states, 1 and 2 , where \(p_{1}=1\) bar, \(V_{1}-1 \mathrm{~m}^{3}, U_{1}=400 \mathrm{~kJ}\) and \(p_{2}-10\) bar, \(V_{2}=0.1 \mathrm{~m}^{3}, U_{2}=450 \mathrm{~kJ}\) : Process A: Constant-volume process from state 1 to a pressure of 10 bar, followed by a constant-pressure process to state \(2 .\) Process B: Process from 1 to 2 during which the pressurevolume relation is \(p V=\) constant. Kinetic and potential effects can be ignored. For each of the processes \(\mathrm{A}\) and \(\mathrm{B}\), (a) sketch the process on \(p-V\) coordinates, (b) evaluate the work, in kJ, and (c) evaluate the heat transfer, in kJ.

A composite plane wall consists of a 3-in.-thick layer of insulation \(\left(\kappa_{\mathrm{s}}=0.029 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot{ }^{\circ} \mathrm{R}\right)\) and a \(0.75\)-in.-thick layer of siding \(\left(\kappa_{\mathrm{s}}=0.058 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot{ }^{\circ} \mathrm{R}\right)\). The inner temperature of the insulation is \(67^{\circ} \mathrm{F}\). The outer temperature of the siding is \(-8^{\circ} \mathrm{F}\). Determine at steady state (a) the temperature at the interface of the two layers, in \({ }^{\circ} \mathrm{F}\), and (b) the rate of heat transfer through the wall in Btu per \(\mathrm{ft}^{2}\) of surface area.

The following table gives data, in \(\mathrm{kJ}\), for a system undergoing a power cycle consisting of four processes in series. Determine, the (a) missing table entries, each in kJ, and (b) the thermal efficiency. $$ \begin{array}{crrr} \text { Process } & \Delta E & Q & W \\ \hline 1-2 & -1200 & 0 & \\ 2-3 & & 800 & \\ 3-4 & & -200 & -200 \\ 4-1 & 400 & & 600 \end{array} $$
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