Problem 4
Question
Why is integration used to find the work done by a variable force?
Step-by-Step Solution
Verified Answer
Answer: Integration is necessary for finding the work done by a variable force because it allows us to sum up the infinitesimal work done at each position, resulting in the total work done by the force. Since the magnitude and direction of a variable force change with respect to its position or other factors, integrating the expression for the infinitesimal work done with respect to the position provides an accurate calculation of the total work done.
1Step 1: Understanding the concept of work done by a variable force
As opposed to a constant force, a variable force has a magnitude and direction that change with respect to its position or other factors. The work done by the force at each point is expressed as dW = F(x) * dx, where dW is the infinitesimal work done, F(x) is the force as a function of position x, and dx is the infinitesimal change in position.
2Step 2: Applying the concept of integration to the work done
To find the total work done by a variable force, we need to sum up the infinitesimal work done at each position. This can be achieved by integrating the expression for dW = F(x) * dx over the given range of positions, say [a, b]. The integration of F(x) * dx with respect to x over the limits [a, b] will give us the total work done by the force. Mathematically, the expression can be written as:
W = \int_a^b F(x)dx
3Step 3: Example problem: Find the work done by a variable force
Let's consider an example problem to understand the process further. Suppose we have a force F(x) = 3x^2 acting on an object in the positive x-direction, where x is the position of the object. We need to find the work done by this force as the object moves from x = 2 m to x = 4 m.
4Step 1: Write down the given force and the limits of integration
The force given is F(x) = 3x^2, and the limits of integration are a = 2 m and b = 4 m.
5Step 2: Integrate the given force function over the specified position range
Now we will integrate the force function F(x) = 3x^2 with respect to x over the limits [a, b], which are [2, 4]:
W = \int_2^4 F(x)dx = \int_2^4 (3x^2)dx
6Step 3: Perform the integration and evaluate the work done
To find the work done, we will solve the integration:
W = \int_2^4 (3x^2)dx = 3 \int_2^4 (x^2)dx = 3 [\frac{1}{3}x^3]_2^4 = x^3 |_2^4 = (4^3 - 2^3)
W = 3 (64 - 8) = 3 * 56 = 168 J
So, the work done by the variable force F(x) = 3x^2 in moving the object from x = 2 m to x = 4 m is 168 J.
Other exercises in this chapter
Problem 4
Suppose \(g\) is positive and differentiable on \([c, d] .\) The curve \(x=g(y)\) on \([c, d]\) is revolved about the \(y\) -axis. Explain how to find the area
View solution Problem 4
What is the inverse function of \(\ln x,\) and what are its domain and range?
View solution Problem 4
The region bounded by the curves \(y=2 x\) and \(y=x^{2}\) is revolved about the \(y\) -axis. Give an integral for the volume of the solid that is generated.
View solution Problem 5
Given the rate of change of a quantity \(Q\) and its initial value \(Q(0)\) explain how to find the value of \(Q\) at a future time \(t \geq 0\)
View solution