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Today, let's talk about how a variable force behaves compared to a constant force when calculating work. Can anyone remind me how we calculate work done by a constant force?
We use the formula W equals force times displacement.
Exactly! Now, if the force varies, how do you think we might handle that?
Maybe we can divide the displacement into small parts where the force is almost constant?
Great thinking! By summing these small segments, we can approximate the total work done. We represent this summation as an integral in calculus. Let's explore this further.
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When we integrate the force function over the displacement, we write it as an integral from an initial position to a final position. Can someone express this mathematically?
It would be W equals the integral of F(x)dx from x_i to x_f.
Perfect! And this integral represents the area under the curve of our force versus displacement graph. Why do you think visualizing this area is helpful?
It helps to see how much work is being done over different parts of displacement!
Exactly! Let's draw a sample graph and calculate the area under the curve to find the total work done by a variable force.
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Can anyone give me an example of a variable force in a real-world scenario?
When I push a shopping cart, the force varies with how much I push, especially when itβs going uphill.
Excellent example! As you push against gravity, the force you exert doesnβt remain constant. Letβs say we draw a graph of force versus distance on this incline.
The area under that curve would show the work done against gravity.
Exactly! We can apply these calculations in engineering, where forces vary continuously.
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To sum up, what are the key points we've discussed regarding the work done by variable forces?
We learned that we can approximate variable forces over small segments.
And we can calculate the total work using definite integrals!
Great! Remember, visualizing force as an area under a curve helps us understand work done more intuitively.
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The section defines work done by a variable force through its graphical representation and integral calculation. It discusses approximating a variable force as constant over small displacements, leading to a cumulative understanding of total work as the area under the force-displacement graph.
In physics, the concept of work done by a variable force diverges from that by a constant force. When a constant force is applied, the work done can be calculated using the equation:
$$W = F imes d imes ext{cos} (ΞΈ)$$
However, in real-world scenarios, forces often vary with position. To analyze these cases, we approximate the variable force by taking small intervals of displacement, where the force can be treated as approximately constant. In this section, we'll go through the method of working it out by calculating the cumulative work done over a series of small displacements.
$$W = ext{lim}{ ext{Ξx} o 0} ext{Ξ£} F(x) ext{Ξx} = ext{β«}{x_i}^{x_f} F(x)dx$$
Understanding the work done by a variable force is crucial in fields such as engineering, physics, and applied sciences where natural forces are often not constant, such as friction, gravity in non-uniform fields, or any forces that vary with position or time.
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A constant force is rare. It is the variable force, which is more commonly encountered. Fig. 5.3 is a plot of a varying force in one dimension. If the displacement βx is small, we can take the force F(x) as approximately constant and the work done is then βW = F(x) βx.
In the context of physics, most forces we deal with in real-life situations are not constant; they change depending on various factors. The term 'variable force' refers to forces that change with position, like the force of a spring when it is compressed or stretched. To calculate work done by such forces, we can approximate the force as constant over small distances. The work done is then the force at that point multiplied by the small change in displacement.
Think of a rubber band. When you stretch it, the force needed to pull it gets greater the more you stretch it. The force applied when just starting to stretch is less than the force needed when it's near its maximum stretch. This is akin to how variable forces work.
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Adding successive rectangular areas in Fig. 5.3(a) we get the total work done as ( )ββ β fi x x x W, where the summation is from the initial position xi to the final position xf.
To find the total work done by a variable force over a distance, you can partition that distance into many small segments. At each segment, you calculate the work done by treating the force during that segment as constant. By summing up all those small works, you obtain the total work done. In a continuous setting, as the number of segments approaches infinity and their width approaches zero, this summation becomes the integral of the force over the displacement.
Imagine sliding a heavy box across a bumpy surface. Initially, it requires less force to move, but as you push it over more bumps, the force requirement increases. If you wanted to calculate the total work done to move the box, you'd consider how much force you needed at each bump along the way, summing them all up to get the total effort.
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If the displacements are allowed to approach zero, then the number of terms in the sum increases without limit, but the sum approaches a definite value equal to the area under the curve in Fig. 5.3(b). Then the work done is W = β«F(x) dx from xi to xf.
As we refine our calculation for the work done by a variable force, we can take smaller and smaller segments. In the limit, as these segments become infinitesimally small, the sum we've devised reverts to the integral of the force function F(x) over the specific displacement from xi to xf. This concept of integral calculus allows us to compute the exact amount of work done over a continuous range, resembling finding the area under a curve.
Consider filling a swimming pool. The rate at which water is traditionally pumped might change depending on the water level (as it fills up). If you were to chart this flow against time, finding how much total water was pumped would involve integrating those changing rates over the time of filling, rather than just taking a simple average.
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Example 5.5: A woman pushes a trunk on a railway platform which has a rough surface. She applies a force of 100 N over a distance of 10 m. Thereafter, she gets progressively tired and her applied force reduces linearly with distance to 50 N. The total distance through which the trunk has been moved is 20 m. Plot the force applied by the woman and the frictional force, which is 50 N versus displacement.
In this example, the woman initially exerts a force of 100 N, but as she tires, this force decreases to 50 N. To calculate the total work done, one would break it into segments based on the varying force. The total work is found by integrating the force over the distance moved, accounting for the frictional force acting against her push as well. The calculation would ultimately show how much work she actually did in moving the trunk, incorporating the constant opposing force of friction.
Think about how running up a hill gets harder. At first, you're able to power up with a lot of energy (applying quite a bit of force), but as you climb higher and tire, you slow down and must exert less force. If you were to map out this force against the distance you run, would find that at different parts of your run, you'd have worked harder in some places than others.
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Key Concepts
Variable Force: This is a force that changes its magnitude over the displacement of the object.
Small Displacements: To calculate work, we can subdivide the overall displacement into small segments (Ξx) where the force can be considered constant.
Definite Integral: The total work done by a variable force is given by the integral of the force function over the displacement. This is expressed mathematically as:
$$W = ext{lim}{ ext{Ξx} o 0} ext{Ξ£} F(x) ext{Ξx} = ext{β«}{x_i}^{x_f} F(x)dx$$
Understanding the work done by a variable force is crucial in fields such as engineering, physics, and applied sciences where natural forces are often not constant, such as friction, gravity in non-uniform fields, or any forces that vary with position or time.
See how the concepts apply in real-world scenarios to understand their practical implications.
Example 1: Calculating work done by a varying force that increases linearly.
Example 2: Determining the work done against gravitational force while moving up an incline.
Example 3: Graphing a variable force versus displacement to find total work done.
Use mnemonics, acronyms, or visual cues to help remember key information more easily.
Force that flexes, moves and flows, Work done by it, the integral shows!
Imagine a ball rolling down a hill; the push it gets from gravity isnβt always the same as it rollsβsometimes fast, sometimes slow. To understand all the work done, we break it down, just like a piece of cake!
W=FΞs: W for Work, F for Force, Ξs for change in position, remember that course!
Review key concepts with flashcards.
Review the Definitions for terms.
Term: Variable Force
Definition:
A force that changes in magnitude or direction over the displacement of an object.
Term: Integral
Definition:
A mathematical operation that represents the area under a curve, used for calculating work done by a variable force.
Term: Cumulative Work
Definition:
The total work done calculated by summing the incremental work done over small displacements.