A variable force acting on a 0.100 -kg particle moving in...

A variable force acting on a 0.100 -kg particle moving in the $x y$ -plane is given by $F(x, y)=\left(x^{2} \hat{x}+y^{2} \hat{y}\right) \mathrm{N},$ where $x$ and $y$ are in meters. Suppose that due to this force, the particle moves from the origin, $O$, to point $S$, with coordinates $(10.0 \mathrm{~m}, 10.0 \mathrm{~m})$. The coordinates of points $P$ and $Q$ are $(0 \mathrm{~m}, 10.0 \mathrm{~m})$ and $(10.0 \mathrm{~m}, 0 \mathrm{~m})$, respectively. Determine the work performed by the force as the particle moves along each of the following paths: a) OPS c) $O S$ e) $O Q S P O$ b) $O Q S$ d) $O P S Q O$

A variable force acting on a 0.100 -kg particle moving in the $x y$ -plane is given by $F(x, y)=\left(x^{2} \hat{x}+y^{2} \hat{y}\right) \mathrm{N},$ where $x$ and $y$ are in meters. Suppose that due to this force, the particle moves from the origin, $O$, to point $S$, with coordinates $(10.0 \mathrm{~m}, 10.0 \mathrm{~m})$. The coordinates of points $P$ and $Q$ are $(0 \mathrm{~m}, 10.0 \mathrm{~m})$ and $(10.0 \mathrm{~m}, 0 \mathrm{~m})$, respectively. Determine the work performed by the force as the particle moves along each of the following paths:
a) OPS
c) $O S$
e) $O Q S P O$
b) $O Q S$
d) $O P S Q O$

Key Concepts

Piecewise Integration of Composite Paths

When a path is broken into several segments, the total work done along the entire path is the sum of the work computed over each segment. In conservative fields, even if the path is composite or involves loops, the net work over any closed path will be zero due to the path independence of work.

Potential Energy Function

For a conservative force field, a corresponding potential energy function exists from which the force can be derived as its gradient. Evaluating the difference in potential energy between the final and initial points gives a direct measure of the work done by the force, irrespective of the path.

Path Independence

Path independence is a direct consequence of a force being conservative. It means that if a particle moves from one point to another, the work done by the force is the same regardless of the route taken. This property simplifies the computation of work in such fields, as calculations can be based solely on endpoint values of the potential function.

Work as a Line Integral

Work done by a force along a path is determined by the line integral of the force over that path. This means integrating the dot product of the force vector and the infinitesimal displacement vector along the trajectory, capturing how much the force is propelling the particle in the direction of motion.

Conservative Force Field

A conservative force field is one in which the work performed moving a particle between any two points depends solely on those endpoints and not on the specific path taken. This implies the existence of a scalar potential function such that the force is the gradient of this potential. The variable force in this context, being expressible as a gradient, exhibits this property.

Transcript

00:01 For this problem on the topic of energy conservation, we have a 0 .1 kilogram particle that is moving in the xy plane under a variable force that is given by fxy, which is x squared x hat plus y squared y hat newton's, where x and y are both in meters.

00:18 The particle moves through this force from the origin to point s with the coordinates 10 meters and 10 meters, and the coordinates of points p and q are 0 and 10 meters.

00:30 And 10 meters and zero respectively.

00:33 We want to find the work done by the force as the particle moves through the following parts.

00:40 Firstly through ops, then through path os, or then through part oqs rather than os, opsqo, and lastly oqspo.

00:54 Now the work done, w is given by the integral over the path from a to b of, d l .f, which is the integral from a to b of x squared dx plus y squared dy.

01:22 Now the equations of the parts are as follows.

01:28 Along o p, x is equal to zero, so d x is along oq, y is equal to 0, so dy is equal to 0.

01:36 Along os, y is equal to x, so dy is equal to d x.

01:41 Along ps, we have y is equal to 10, so dy is equal to 0.

01:45 And along qs, x is equal to 10, so d x is equal to 0.

01:49 So we can calculate the work done over each of these parts.

01:54 So over part ops, we have the work done to be the integral from o to p, of x squared d x plus y squared d y d y plus the integral from p to s of x squared d x plus y squared d y and these integrals become the integral from 0 to 10 of y squared d .y plus the integral from 0 to 10 of x squared d x.

02:46 And so this is equal to 1 over 3, y cubed evaluated from 0 to 10, plus 1 over 3 x cubed evaluated from x going from 0 to 10.

02:59 And so this is equal to 1 over 3 x cubed, evaluated from x going from 0 to 10...

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