Work Done by the Applied Force

Oct 13, 2025 by miodeya.com 108 views

Introduction

In physics, work is a fundamental concept that describes the transfer of energy from one object to another through a force applied over a distance. The work done on an object is a crucial aspect of understanding various physical phenomena, including motion, energy, and power. In this article, we will delve into the concept of work done on an object, focusing on a specific scenario where a crate is pushed on a flat surface with a force at an angle. We will calculate the total amount of work done, taking into account the frictional force exerted back onto the crate.

Understanding Work Done

Work done on an object is defined as the product of the force applied and the displacement of the object in the direction of the force. Mathematically, it can be represented as:

W = \vec{F} \cdot \vec{d}

where $W$ is the work done, $\vec{F}$ is the force applied, and $\vec{d}$ is the displacement of the object.

Calculating Work Done

In our scenario, the crate is pushed with a force of 88.4 N at an angle of $65.5^{\circ}$ over a distance of 11.9 m. To calculate the work done, we need to find the component of the force in the direction of the displacement. This can be done using trigonometry:

F_{\text{parallel}} = F \cos \theta

where $F_{\text{parallel}}$ is the component of the force in the direction of the displacement, $F$ is the magnitude of the force, and $\theta$ is the angle between the force and the displacement.

Plugging in the values, we get:

F_{\text{parallel}} = 88.4 \cos 65.5^{\circ} = 88.4 \times 0.423 = 37.3 \text{ N}

Now, we can calculate the work done using the formula:

W = F_{\text{parallel}} d

where $d$ is the displacement of the object.

Plugging in the values, we get:

W = 37.3 \times 11.9 = 443.57 \text{ J}

Frictional Force

However, we need to take into account the frictional force exerted back onto the crate. The frictional force is given as 23.2 N. To calculate the work done by the frictional force, we need to find the component of the frictional force in the direction of the displacement. Since the frictional force is perpendicular to the displacement, the component of the frictional force in the direction of the displacement is zero.

Therefore, the work done by the frictional force is zero.

Total Work Done

The total work done is the sum of the work done by the applied force and the work done by the frictional force:

W_{\text{total}} = W + W_{\text{friction}}

where $W_{\text{total}}$ is the total work done, $W$ is the work done by the applied force, and $W_{\text{friction}}$ is the work done by the frictional force.

Plugging in the values, we get:

W_{\text{total}} = 443.57 + 0 = 443.57 \text{ J}

Conclusion

In conclusion, the total amount of work done on the crate is 443.57 J. This is the energy transferred to the crate through the applied force, taking into account the frictional force exerted back onto the crate. The work done is a crucial aspect of understanding various physical phenomena, including motion, energy, and power.

References

[1] Halliday, D., Resnick, R., & Walker, J. (2013). Fundamentals of Physics. John Wiley & Sons.
[2] Serway, R. A., & Jewett, J. W. (2018). Physics for Scientists and Engineers. Cengage Learning.

Mathematical Derivations

Work Done by the Applied Force

The work done by the applied force is given by:

W = \vec{F} \cdot \vec{d}

where $W$ is the work done, $\vec{F}$ is the force applied, and $\vec{d}$ is the displacement of the object.

To calculate the work done, we need to find the component of the force in the direction of the displacement. This can be done using trigonometry:

F_{\text{parallel}} = F \cos \theta

where $F_{\text{parallel}}$ is the component of the force in the direction of the displacement, $F$ is the magnitude of the force, and $\theta$ is the angle between the force and the displacement.

Plugging in the values, we get:

F_{\text{parallel}} = 88.4 \cos 65.5^{\circ} = 88.4 \times 0.423 = 37.3 \text{ N}

Now, we can calculate the work done using the formula:

W = F_{\text{parallel}} d

where $d$ is the displacement of the object.

Plugging in the values, we get:

W = 37.3 \times 11.9 = 443.57 \text{ J}

Work Done by the Frictional Force

The frictional force is given as 23.2 N. To calculate the work done by the frictional force, we need to find the component of the frictional force in the direction of the displacement. Since the frictional force is perpendicular to the displacement, the component of the frictional force in the direction of the displacement is zero.

Therefore, the work done by the frictional force is zero.

Total Work Done

The total work done is the sum of the work done by the applied force and the work done by the frictional force:

W_{\text{total}} = W + W_{\text{friction}}

where $W_{\text{total}}$ is the total work done, $W$ is the work done by the applied force, and $W_{\text{friction}}$ is the work done by the frictional force.

Plugging in the values, we get:

W_{\text{total}} = 443.57 + 0 = 443.57 \text{ J}$$<br/> **Work Done on an Object: A Comprehensive Analysis** =====================================================

Q&A: Frequently Asked Questions

Q: What is work done on an object?

A: Work done on an object is the transfer of energy from one object to another through a force applied over a distance. It is a measure of the energy expended in moving an object from one point to another.

Q: How is work done calculated?

A: Work done is calculated using the formula:

W = \vec{F} \cdot \vec{d} </span></p> <p>where <span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>W</mi></mrow><annotation encoding="application/x-tex">W</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.6833em;"></span><span class="mord mathnormal" style="margin-right:0.13889em;">W</span></span></span></span> is the work done, <span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mover accent="true"><mi>F</mi><mo>⃗</mo></mover></mrow><annotation encoding="application/x-tex">\vec{F}</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.9663em;"></span><span class="mord accent"><span class="vlist-t"><span class="vlist-r"><span class="vlist" style="height:0.9663em;"><span style="top:-3em;"><span class="pstrut" style="height:3em;"></span><span class="mord mathnormal" style="margin-right:0.13889em;">F</span></span><span style="top:-3.2523em;"><span class="pstrut" style="height:3em;"></span><span class="accent-body" style="left:-0.1522em;"><span class="overlay" style="height:0.714em;width:0.471em;"><svg xmlns="http://www.w3.org/2000/svg" width="0.471em" height="0.714em" style="width:0.471em" viewBox="0 0 471 714" preserveAspectRatio="xMinYMin"><path d="M377 20c0-5.333 1.833-10 5.5-14S391 0 397 0c4.667 0 8.667 1.667 12 5 3.333 2.667 6.667 9 10 19 6.667 24.667 20.333 43.667 41 57 7.333 4.667 11 10.667 11 18 0 6-1 10-3 12s-6.667 5-14 9c-28.667 14.667-53.667 35.667-75 63 -1.333 1.333-3.167 3.5-5.5 6.5s-4 4.833-5 5.5c-1 .667-2.5 1.333-4.5 2s-4.333 1 -7 1c-4.667 0-9.167-1.833-13.5-5.5S337 184 337 178c0-12.667 15.667-32.333 47-59 H213l-171-1c-8.667-6-13-12.333-13-19 0-4.667 4.333-11.333 13-20h359 c-16-25.333-24-45-24-59z"/></svg></span></span></span></span></span></span></span></span></span></span> is the force applied, and <span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mover accent="true"><mi>d</mi><mo>⃗</mo></mover></mrow><annotation encoding="application/x-tex">\vec{d}</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.9774em;"></span><span class="mord accent"><span class="vlist-t"><span class="vlist-r"><span class="vlist" style="height:0.9774em;"><span style="top:-3em;"><span class="pstrut" style="height:3em;"></span><span class="mord mathnormal">d</span></span><span style="top:-3.2634em;"><span class="pstrut" style="height:3em;"></span><span class="accent-body" style="left:-0.0688em;"><span class="overlay" style="height:0.714em;width:0.471em;"><svg xmlns="http://www.w3.org/2000/svg" width="0.471em" height="0.714em" style="width:0.471em" viewBox="0 0 471 714" preserveAspectRatio="xMinYMin"><path d="M377 20c0-5.333 1.833-10 5.5-14S391 0 397 0c4.667 0 8.667 1.667 12 5 3.333 2.667 6.667 9 10 19 6.667 24.667 20.333 43.667 41 57 7.333 4.667 11 10.667 11 18 0 6-1 10-3 12s-6.667 5-14 9c-28.667 14.667-53.667 35.667-75 63 -1.333 1.333-3.167 3.5-5.5 6.5s-4 4.833-5 5.5c-1 .667-2.5 1.333-4.5 2s-4.333 1 -7 1c-4.667 0-9.167-1.833-13.5-5.5S337 184 337 178c0-12.667 15.667-32.333 47-59 H213l-171-1c-8.667-6-13-12.333-13-19 0-4.667 4.333-11.333 13-20h359 c-16-25.333-24-45-24-59z"/></svg></span></span></span></span></span></span></span></span></span></span> is the displacement of the object.</p> <h2><strong>Q: What is the difference between work done and energy?</strong></h2> <p>A: Work done is the transfer of energy from one object to another, while energy is the ability to do work. Energy can be transferred from one object to another through various means, including work done.</p> <h2><strong>Q: Can work done be negative?</strong></h2> <p>A: Yes, work done can be negative. This occurs when the force applied is in the opposite direction of the displacement, resulting in a decrease in energy.</p> <h2><strong>Q: What is the unit of work done?</strong></h2> <p>A: The unit of work done is the joule (J).</p> <h2><strong>Q: Can work done be zero?</strong></h2> <p>A: Yes, work done can be zero. This occurs when the force applied is perpendicular to the displacement, resulting in no energy transfer.</p> <h2><strong>Q: What is the relationship between work done and friction?</strong></h2> <p>A: Friction can affect the work done on an object. When friction is present, it can reduce the work done by opposing the motion of the object.</p> <h2><strong>Q: Can work done be calculated using other formulas?</strong></h2> <p>A: Yes, work done can be calculated using other formulas, such as:</p> <p class='katex-block'><span class="katex-display"><span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML" display="block"><semantics><mrow><mi>W</mi><mo>=</mo><mi>F</mi><mi>d</mi><mi>cos</mi><mo>⁡</mo><mi>θ</mi></mrow><annotation encoding="application/x-tex">W = F d \cos \theta </annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.6833em;"></span><span class="mord mathnormal" style="margin-right:0.13889em;">W</span><span class="mspace" style="margin-right:0.2778em;"></span><span class="mrel">=</span><span class="mspace" style="margin-right:0.2778em;"></span></span><span class="base"><span class="strut" style="height:0.6944em;"></span><span class="mord mathnormal" style="margin-right:0.13889em;">F</span><span class="mord mathnormal">d</span><span class="mspace" style="margin-right:0.1667em;"></span><span class="mop">cos</span><span class="mspace" style="margin-right:0.1667em;"></span><span class="mord mathnormal" style="margin-right:0.02778em;">θ</span></span></span></span></span></p> <p>where <span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>W</mi></mrow><annotation encoding="application/x-tex">W</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.6833em;"></span><span class="mord mathnormal" style="margin-right:0.13889em;">W</span></span></span></span> is the work done, <span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>F</mi></mrow><annotation encoding="application/x-tex">F</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.6833em;"></span><span class="mord mathnormal" style="margin-right:0.13889em;">F</span></span></span></span> is the force applied, <span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>d</mi></mrow><annotation encoding="application/x-tex">d</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.6944em;"></span><span class="mord mathnormal">d</span></span></span></span> is the displacement of the object, and <span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>θ</mi></mrow><annotation encoding="application/x-tex">\theta</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.6944em;"></span><span class="mord mathnormal" style="margin-right:0.02778em;">θ</span></span></span></span> is the angle between the force and the displacement.</p> <h2><strong>Q: What is the significance of work done in real-world applications?</strong></h2> <p>A: Work done is significant in various real-world applications, including:</p> <ul> <li><strong>Mechanical engineering</strong>: Work done is used to calculate the energy required to move objects, such as machines and mechanisms.</li> <li><strong>Aerospace engineering</strong>: Work done is used to calculate the energy required to launch spacecraft and satellites.</li> <li><strong>Biomechanics</strong>: Work done is used to calculate the energy required to move the human body, such as during exercise and physical activity.</li> </ul> <h2><strong>Q: Can work done be calculated using numerical methods?</strong></h2> <p>A: Yes, work done can be calculated using numerical methods, such as:</p> <ul> <li><strong>Numerical integration</strong>: Work done can be calculated using numerical integration methods, such as the trapezoidal rule and Simpson's rule.</li> <li><strong>Finite element methods</strong>: Work done can be calculated using finite element methods, such as the finite element method (FEM) and the boundary element method (BEM).</li> </ul> <h2><strong>Conclusion</strong></h2> <p>In conclusion, work done is a fundamental concept in physics that describes the transfer of energy from one object to another through a force applied over a distance. It is a measure of the energy expended in moving an object from one point to another. Work done can be calculated using various formulas and numerical methods, and it has significant applications in various fields, including mechanical engineering, aerospace engineering, and biomechanics.</p>