Work Done By a Force - Incline Planes & Dot Product Formula - Physics

The Organic Chemistry Tutor

28 Sept 202317:29

EducationalLearning

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TLDRThis video tutorial explains how to calculate work done by a force using the dot product formula. It demonstrates the process with examples, starting with a basic scenario involving a force of 300 Newtons at a 40-degree angle, displacing a block 5 meters horizontally, resulting in 1149 Joules of work. The video further illustrates the calculation using vector components for both force and displacement, and applies the method to more complex situations like pulling a block up an incline. The explanation clarifies the concepts of positive and negative work, emphasizing the importance of the direction of force relative to displacement.

Takeaways

📚 The script introduces a method to calculate work done by a force using the dot product formula.
🔢 The basic formula for work is the magnitude of the force times the magnitude of the object's displacement times the cosine of the angle between them.
📏 An example is given with a force of 300 Newtons at a 40-degree angle, displacing a block 5 meters horizontally, resulting in 1149 Joules of work.
📐 The dot product formula for work involves multiplying the components of the force vector and the displacement vector and summing the products.
📉 For a two-dimensional vector, the formula includes the X and Y components, and for three dimensions, it would also include the Z components.
📚 The script demonstrates converting force and displacement vectors into their component forms to use in the dot product formula.
📝 The force vector components are calculated using the magnitude of the force and the angle, resulting in X and Y components.
📍 The displacement vector, when parallel to the x-axis, simplifies to just the magnitude times the cosine of the angle (which is 1 for 0 degrees).
🔄 The script shows how to calculate work done in both the X and Y directions separately and then totals them to get the overall work done.
🔢 Another example is given with a force applied at an angle to an inclined plane, demonstrating both the direct formula and the dot product method.
📉 The script concludes with an exercise for the viewer to calculate work done with given force and displacement vectors in component form.
👉 The difference between positive and negative work is explained based on the direction of the force relative to the displacement vector.

Q & A

What is the basic formula to calculate the work done by a force?
-The basic formula to calculate the work done by a force is the magnitude of the force (F) times the magnitude of the object's displacement (D) times the cosine of the angle (Theta) between the force and displacement vectors, which can be represented as W = F * D * cos(Theta).
What is the dot product formula used to calculate work done by a force?
-The dot product formula for calculating work done by a force is W = F · D, where F is the force vector and D is the displacement vector. It is calculated by multiplying the corresponding components of the force and displacement vectors and summing the results.
How do you convert a force vector into its component form?
-A force vector is converted into its component form by multiplying the magnitude of the force by the cosine of the angle for the x-component and by the sine of the angle for the y-component, using the unit vectors i and j, respectively.
What is the significance of the angle between the force and displacement vectors in calculating work?
-The angle between the force and displacement vectors is significant because it determines the component of the force that is doing work along the direction of displacement. The work done is maximum when the angle is 0 degrees (force and displacement are in the same direction) and zero when the angle is 90 degrees (force is perpendicular to displacement).
What is the result of the first example calculation in the script, where a force of 300 Newtons is applied at a 40-degree angle to displace an object 5 meters horizontally?
-The result of the first example calculation is 1149 Joules, which is obtained by plugging the values into the formula W = F * D * cos(Theta), where F = 300 N, D = 5 m, and Theta = 40 degrees.
How does the dot product of vectors relate to the work done in the x and y directions separately?
-The dot product of vectors allows you to calculate the work done separately in the x and y directions by multiplying the x-components of the force and displacement vectors for the x-direction, and the y-components for the y-direction. The total work done is the sum of these two directional works.
What is the difference between positive and negative work?
-Positive work is done when the force and displacement vectors are in the same direction, indicating that the force is aiding the motion. Negative work is done when the force is in the opposite direction to the displacement, indicating that the force is opposing the motion. If the force is perpendicular to the displacement, no work is done.
How can you calculate the work done using the dot product formula if you have a three-dimensional force and displacement vector?
-In a three-dimensional case, you would calculate the work done using the dot product formula by including the z-components of both the force and displacement vectors, in addition to the x and y components. The formula would be W = Fx * Dx + Fy * Dy + Fz * Dz.
What does the result of the second example calculation in the script, where a force of 200 Newtons is applied at a 30-degree angle to an incline of 20 degrees for a displacement of 12 meters, indicate?
-The result of the second example calculation is approximately 2078 Joules, indicating the total work done by the force in pulling the block up the incline, which includes work done in both the x and y directions.
How can you interpret the components of work done in the x and y directions from the dot product calculation?
-The components of work done in the x and y directions from the dot product calculation represent the work done along each axis separately. The total work done is the sum of these components, which can be positive or negative depending on the direction of the force relative to the displacement.

Outlines

00:00

🔍 Calculating Work Done Using Dot Product

This paragraph introduces the concept of calculating work done by a force using the dot product formula. It explains the basic formula for work done as the magnitude of the force times the magnitude of the object's displacement times the cosine of the angle between them. An example is given with a force of 300 Newtons at a 40-degree angle, displacing a block 5 meters to the right, resulting in 1149 Joules of work. The paragraph then transitions into explaining how to calculate the same using the dot product of force and displacement vectors, emphasizing the need to convert these vectors into their component forms for the calculation.

05:01

📚 Work Done Calculation with Dot Product Formula

The second paragraph delves deeper into calculating work done using the dot product formula. It provides a step-by-step guide on converting force and displacement vectors into their component forms and then calculating the dot product to find the work done. An example with a block on an incline and a force applied at a 30-degree angle to the incline is used to illustrate the process. The paragraph concludes by comparing the result obtained through the dot product method with the simpler method, both yielding 2078 Joules of work.

10:03

📈 Breakdown of Work Done in X and Y Directions

In this paragraph, the script explains how to calculate the work done in both the X and Y directions when force and displacement vectors are involved. It uses the dot product of the components of the force and displacement vectors to find the work done separately in each direction. The example provided shows calculations for a force vector with both X and Y components and a displacement vector on an incline, resulting in 1449 Joules of work in the X direction and 628.80 Joules in the Y direction, summing up to a total of 2078 Joules.

15:05

🔄 Understanding Positive and Negative Work

The final paragraph discusses the concept of positive and negative work. It explains that positive work is done when the force and displacement vectors are in the same direction, while negative work is done when they are in opposite directions. The paragraph uses an example with given force and displacement vectors in component form to demonstrate the calculation, resulting in negative work done. It concludes by clarifying the conditions under which positive or negative work is considered and invites viewers to explore more practice problems on work and energy through provided links.

Mindmap

Keywords

💡Work Done

Work done is a measure of energy transfer that occurs when a force causes a displacement of an object. In the video, it is the central concept being explained, with the formula for calculating it given as the product of the magnitude of the force, the magnitude of the object's displacement, and the cosine of the angle between the force and displacement vectors. The script provides examples of calculating work done using both the direct formula and the dot product method.

💡Dot Product

The dot product is a mathematical operation that takes two vectors and returns a scalar. It is used in the script to calculate the work done by a force. The explanation involves multiplying the corresponding components of the force and displacement vectors and summing the results to find the total work done. This method is demonstrated with examples in the video.

💡Force Vector

A force vector is a vector quantity that represents both the magnitude and the direction of a force acting on an object. In the script, the force vector is broken down into its x and y components using trigonometric functions based on the angle it makes with the horizontal. This is crucial for determining the work done as the force vector's components are used in the dot product calculation.

💡Displacement Vector

The displacement vector represents the change in position of an object, with both magnitude (how far) and direction. In the video, the displacement vector is used to calculate work done, and its components are determined based on the direction of movement. It is essential in the dot product formula to find the work done by a force.

💡Magnitude

Magnitude refers to the size or length of a vector, which in the context of the video, is used to describe the strength of the force or the distance of the displacement. The magnitude of the force and displacement vectors is necessary to calculate the work done using the formula provided in the script.

💡Cosine Theta

Cosine Theta is used in the formula for work done to account for the angle between the force and displacement vectors. In the script, it is part of the initial formula for work done (F * D * cos(Theta)), where Theta is the angle between the force and displacement vectors. The cosine of this angle determines how much of the force contributes to the work done.

💡Component Form

Component form refers to expressing a vector in terms of its x, y, and sometimes z components. In the video, both the force and displacement vectors are converted into their component form to facilitate the calculation of the dot product, which is then used to find the work done.

💡Unit Vector

A unit vector is a vector with a magnitude of one, typically used to represent direction in a coordinate system. In the script, unit vectors i and j are used to express the force and displacement vectors in component form along the x and y axes, respectively.

💡Angle

The angle in the context of the video is the angle between the force vector and the displacement vector. It is a critical factor in determining the work done, as it affects the value of cosine Theta in the work formula. The script discusses calculating work done with different angles to demonstrate its impact.

💡Positive and Negative Work

Positive and negative work are terms used to describe the effect of the direction of the force relative to the displacement. In the script, it is explained that if the force and displacement are in the same direction, the work is positive, indicating energy is transferred in the direction of displacement. Conversely, if they are in opposite directions, the work is negative, indicating energy is transferred opposite to the displacement.

💡Three-Dimensional Vector

A three-dimensional vector has components in the x, y, and z directions. While the primary focus of the video is on two-dimensional vectors, the script briefly mentions that for three-dimensional vectors, an additional component (fz * dz) would be included in the dot product formula to calculate work done.

Highlights

Introduction to calculating work done by a force using the dot product formula.

Explanation of the basic formula for work done: magnitude of force times displacement times cosine of the angle between them.

Example calculation with a force of 300 Newtons at a 40-degree angle and a 5-meter displacement.

Result of the example calculation: 1149 Joules as the simplest way to find work done.

Introduction to calculating work using the dot product of force and displacement vectors.

Process of converting force and displacement vectors into their component forms for the dot product calculation.

Component breakdown of a force vector with a magnitude of 300 Newtons at a 40-degree angle.

Component breakdown of a displacement vector with a magnitude of 5 meters parallel to the x-axis.

Calculation of work done using the dot product formula, resulting in 11.49 Joules.

Introduction to a second example involving an inclined plane with a force applied at a 30-degree angle above the incline.

Calculation of work done pulling a block up an incline using both the basic formula and the dot product formula.

Component form of the force vector for the inclined plane example with a magnitude of 200 Newtons at a 50-degree angle.

Component form of the displacement vector for the inclined plane example with a magnitude of 12 meters at a 20-degree angle.

Dot product calculation for the inclined plane example, resulting in approximately 2078 Joules.

Explanation of the difference between positive and negative work done by a force.

Final example with given force and displacement vectors in component form and the calculation of work done.

Result of the final example showing negative work done, indicating the force is mostly opposite to the object's displacement.

Summary of the conditions for positive, negative, and zero work done by a force relative to displacement.

Invitation to check out more practice problems on work and energy in the description section.

Transcripts

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