Irodov 1.118: Only the Endpoints Ever Mattered – Amit & Deepali Sinha Foundation

Problem Irodov 1.118

A particle moves along some unstated curved path in the $xy$ plane from a point $1$ , located by the position vector $\vect r_1=\vect i+2\vect j$ , to a point $2$ , located by the position vector $\vect r_2=2\vect i-3\vect j$ (components in SI units). Several forces act on the particle during the trip, and among them is one constant force $\vect F=3\vect i+4\vect j$ (also in SI units). Find the work done by $\vect F$ alone as the particle makes this trip.

The position vectors r_1, r_2 of the two points and the net displacement r_2- r_1 between them, dashed. F is drawn schematically at point 2: a force and a position carry different units and cannot share one length scale. — The position vectors $\vect r_1$ , $\vect r_2$ of the two points and the net displacement $\vect r_2-\vect r_1$ between them, dashed. $\vect F$ is drawn schematically at point $2$ : a force and a position carry different units and cannot share one length scale.

Given

$\vect r_1=\vect i+2\vect j$
$\vect r_2=2\vect i-3\vect j$
$\vect F=3\vect i+4\vect j$ , all in SI units

Find

The work

A

done by

\vect F

as the particle moves from point

1

to point

2

Solution

Step 1Work of a constant force needs only the endpoints

The problem gives $\vect F$ as one fixed vector, not a function of position, even though other unnamed forces act on the particle meanwhile. When a force does not change along the path, it comes out of the line integral that defines work, so the integral collapses to the two endpoints regardless of the trajectory shape between them.

\begin{align*}A = \int_1^2 \vect F\cdot d\vect r = \vect F\cdot\int_1^2 d\vect r.\end{align*}

A=\vect F\cdot(\vect r_2-\vect r_1)

(1)

Step 2The net displacement

Subtracting the two position vectors component by component gives the displacement (1) needs.

\begin{align*}\vect r_2-\vect r_1 = (2\vect i-3\vect j)-(\vect i+2\vect j) = \vect i - 5\vect j.\end{align*}

Step 3Evaluate the dot product

Using (1) with $\vect F=3\vect i+4\vect j$ and $\vect r_2-\vect r_1=\vect i-5\vect j$ :

\begin{align*}A = (3)(1) + (4)(-5) = 3 - 20.\end{align*}

A=-17\ \text{J}

(2)

Check

The $y$ component dominates: $F_y=4$ multiplies a displacement of $-5$ along $y$ , contributing $-20$ J, against only $+3$ J from the $x$ component, so a net negative, energy removing, work is exactly what the numbers give.

Remark

A constant force is the simplest case of a potential field: taking $U(\vect r)=-\vect F\cdot\vect r$ gives $\vect F=-\nabla U$ automatically, so the shortcut used here, that work needs only the endpoints, is really the general path independence of a conservative field rather than a special trick tied to $\vect F$ being constant; that endpoint-only property survives for any conservative force, position dependent or not. What is special to a constant $\vect F$ is only that the endpoint expression takes the simple form $\vect F\cdot(\vect r_2-\vect r_1)$ ; the collapse fails only for a non-conservative force.