Integration Intuition for work between x1 and x2

[Joshb60796]

Homework Statement

A force F=bx³ acts in the x direction, where the value of b is 3.7N/m³. How much work is done by this force in moving an object from x=0.00m to x=2.6m?

Homework Equations

W=F*D
∫uⁿdu = (u⁽ⁿ⁺¹⁾/(n+1))+C

The Attempt at a Solution

I know from previous problems that I'm going to integrate to find the area under the curve bx³ from 0 to 2.6 but I don't have the intuition as to why. I don't really understand why I'm doing this besides the face that I see a function raised to a power and not much more information is given. I'm following a pattern based on a pattern I see. I get that the force is changing in the y coordinate and that the x coordinate I'm worried about goes from 0 to 2.6 and I'm finding the area using integration but is the force a jerk because it's raised to the 3rd power, would it be acceleration if raised to the second power? I remember doing derivatives and thinking this. How did they come up with b? How did b come to be? What measuring device gives a reading like 3.7N/m³? I'm not even sure I'm asking the right questions but I want to fully understand this.

 

[rock.freak667]

The Attempt at a Solution

I know from previous problems that I'm going to integrate to find the area under the curve bx³ from 0 to 2.6 but I don't have the intuition as to why.

If you consider a force F moving its point of application an infinitesimal distance 'dx', then the work done within in this distance is dW = F dx

For the work done over the entire distance from x₂ to x₁

\int_{0} ^W 1 dW = \int_{x_1} ^{x_2} F dx \Rightarrow W = \int_{x_1} ^{x_2} F dx

Graphically, this represents the area under the force-distance curve.

I don't really understand why I'm doing this besides the face that I see a function raised to a power and not much more information is given. I'm following a pattern based on a pattern I see. I get that the force is changing in the y coordinate and that the x coordinate I'm worried about goes from 0 to 2.6 and I'm finding the area using integration but is the force a jerk because it's raised to the 3rd power, would it be acceleration if raised to the second power? I remember doing derivatives and thinking this. How did they come up with b? How did b come to be? What measuring device gives a reading like 3.7N/m³? I'm not even sure I'm asking the right questions but I want to fully understand this.

What could happen to get such a formula is that based on measuring force with distance, you could find that F∝x³ such that F= bx³. If you plot a graph of F against x³, the gradient would be 'b'.

To get what units 'b' would have, the units of the product of 'bx³' would need to be Newtons (N); x is in m.

so N = b*m³ such that b = N/m³. So you would not always have an instrument measuring something in N/m³ but you might be able to figure it out based on other data.

 

[Joshb60796]

Thank you sir for your insight. What was that symbol you used after F? I've never seen it before. Also, what is a gradient exactly? Is it like a coefficient that changes? I've only had up to Calculus 2 so far.

 

[rock.freak667]

Thank you sir for your insight. What was that symbol you used after F? I've never seen it before. Also, what is a gradient exactly? Is it like a coefficient that changes? I've only had up to Calculus 2 so far.

The ∝ ? I am not sure why it appears so small but it is the symbol for 'directly proportional to' so as one increases, the other increases by a set amount.

Gradient refers to a rate of change. In the straight line equation y=mx+c, the gradient would be equal to m. Such that the change in the distance y (denoted by Δy) divided by the change in the distance x (Δx) is equal to m.

Δy/Δx = m

For more information on a basic method on how gradient or slope is calculated, you can read here.

 

[Joshb60796]

oh ok, I didn't realize that gradient was synonymous to slope. Thank you for the explanation.