Work Done by a General Variable Force — Question 6

Question 6

Two integral measurements identify a linear force

A force is known to have the form $F_x=a+bx$. Its work from 0 to 1 m is 5 J; from 1 to 2 m it is 1 J. Treat the data as exact.

1. Find a and b, including their units.

2. Locate the force zero.

3. Find work from 0 to 3 m and explain how it can be smaller than work from 0 to 2 m.

Question 6 - Solution

Two integral measurements identify a linear force

(a) Translate each interval into an equation.

The governing relation is $$\boxed{W_{p\to q}=a(q-p)+\tfrac12b(q^2-p^2)}$$

$$a(1\,\mathrm m)+\tfrac12b(1\,\mathrm m^2)=5\ \mathrm J,\qquad a(1\,\mathrm m)+\tfrac32b(1\,\mathrm m^2)=1\ \mathrm J.$$ Subtraction gives $$b=-4\ \mathrm{N/m}$$ and then $$a=7\ \mathrm N.$$

(b) Set the reconstructed force to zero.

$7-4x=0$ at x=1.75 m. The force opposes rightward motion beyond this position.

(c) Integrate over the longer interval.

$$W_{0\to3}=[7x-2x^2]_0^3=3\ \mathrm J.$$ Work through 2 m is 6 J; the additional interval contributes $-3\ \mathrm J$. A longer displacement need not increase signed work.