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Chapter 6: Problem 4

Why must integration be used to find the work done by a variable force?

Short Answer

Expert verified
Answer: Integration is necessary to find the work done by a variable force because the force is not constant over the entire displacement. It allows us to break down the problem into small increments of distance, considering the force as constant within each increment, and then sum up all these small amounts of work. This provides a more accurate representation of the total work done by a variable force.

Step by step solution

01

Understanding work done by a constant force

When dealing with a constant force, the work done (W) can be found using the formula W = F * d * cos(theta), where F is the force, d is the displacement, and theta is the angle between the force and displacement vectors. However, this formula only works for constant forces.
02

Explaining why variable forces need a different approach

In cases where the force is variable, we cannot use the above formula directly because the force changes over the distance it is applied. Instead, we need to break down the problem into small increments of displacement (螖x) over which the force can be considered constant. For each increment, we can then compute the small amount of work (螖W) done using the formula mentioned in Step 1.
03

Introducing integration

To find the total work done by a variable force, we need to sum up all the small amounts of work done (螖W) over each increment of displacement (螖x). As the intervals get smaller and more precise (approaching infinitesimally small), the process of summing these incremental works approaches integration.
04

Writing the integral for work done

In cases where the force F is variable, we can find the total work done by integrating the variable force over the range of displacement. Using the integral notation, we can write it as: W = 鈭�(F(x) * cos(theta) dx) This integral runs from the initial displacement (x1) to the final displacement (x2), where F(x) is the variable force as a function of x, and theta is the angle between the force and displacement vectors at each point.
05

Conclusion

In conclusion, integration must be used to find the work done by a variable force because the force is not constant over the entire displacement. Integration allows us to compute the work done by dividing the problem into small increments of distance, considering the force as constant within each increment, and then summing up all these small amounts of work.

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Key Concepts

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

Work Done
When we talk about the work done, it essentially means the effort exerted when applying a force to move an object over a displacement. For a constant force, this is straightforward, as we can use the formula:
  • \( W = F \cdot d \cdot \cos(\theta) \)
where \( F \) is the force applied, \( d \) is the displacement, and \( \theta \) is the angle between the force and direction of movement.

When the force is constant, the work done remains the same throughout the motion. However, in reality, forces can vary based on many factors such as speed, friction, or altitude. Our goal: find the total work when dealing with variable forces.
Integration
Integration is a crucial mathematical tool used when dealing with variable forces. Integration helps us add up small increments and changes, allowing us to calculate the total effect.

Instead of tackling the entire problem with one formula, integration takes many tiny slices of the problem, adding them together for an accurate total. By integrating the forces acting across the displacement of an object, we can accurately compute the work done, irrespective of variations.

In simple terms, integration allows us to sum up infinite small parts to form a wholesome result, ideal for non-uniform forces.
Displacement
The term displacement refers to the change in position of an object from one point to another. It is a vector quantity, which means both magnitude and direction are considered.

When applying a force to displace an object, the displacement becomes an essential part of the work calculation. For variable forces, we consider tiny increments of displacement over which the force is quite possibly constant.

By breaking down the displacement into smaller segments, it allows for easier calculation and integration of variable forces to determine the overall work done. In terms of integration, we assess each tiny piece of displacement separately.
Integral Calculus
Integral calculus delves into the idea of finding areas under curves. In the context of work done, this means calculating the cumulative work over the path of the object.

Imagine a graph where force varies along the x-axis (displacement). The area under this curve provides the work done by the force. Calculating this area for variable forces is possible through the integral calculus, where we write:
  • \( W = \int F(x) \cdot \cos(\theta) \, dx \)
The integral sums these small increments of work across the chosen range, from starting point \(x1\) to endpoint \(x2\).

Overall, integral calculus is vital for evaluating total work done in scenarios with non-constant forces.

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Most popular questions from this chapter

Suppose a force of \(30 \mathrm{N}\) is required to stretch and hold a spring \(0.2 \mathrm{m}\) from its equilibrium position. a. Assuming the spring obeys Hooke's law, find the spring constant \(k\) b. How much work is required to compress the spring \(0.4 \mathrm{m}\) from its equilibrium position? c. How much work is required to stretch the spring \(0.3 \mathrm{m}\) from its equilibrium position? d. How much additional work is required to stretch the spring \(0.2 \mathrm{m}\) if it has already been stretched \(0.2 \mathrm{m}\) from its equilibrium position?

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