Work Done by a General Variable Force — Question 1

Question 1

Signed areas can cancel even when force never stays zero

A particle moves in the positive x direction from 0 to 4 m. The plotted force is linear between the points $(0,0)$, $(1,6)$, $(2,0)$, $(3,-3)$, and $(4,0)$, with x in meters and force in newtons.

1. Find work on the first and second halves of the interval.

2. Find total work and the displacement-averaged force.

3. If this is the only force doing work and initial kinetic energy is 1 J, is the entire route accessible?

Question 1 - Solution

Signed areas can cancel even when force never stays zero

(a) Use signed triangular areas.

The governing relation is $$\boxed{W_{a\to b}=\int_a^bF_x(x)\,dx\quad\text{(signed area)}}$$

$$W_{0\to2}=\tfrac12(2)(6)=6\ \mathrm J.$$

$$W_{2\to4}=-\tfrac12(2)(3)=-3\ \mathrm J.$$

(b) Add before dividing by displacement.

$$W=3\ \mathrm J,\qquad \overline F_x=3/4=0.75\ \mathrm N.$$ The mean of the maximum and minimum forces is not a displacement average for this graph.

(c) Check intermediate kinetic energy.

K rises from 1 to 7 J, then falls to 4 J. It stays positive everywhere, so a particle initially moving right does not turn before x=4 m.