Work — Question 4

Question 4

A force schedule matters even when displacement cancels

An actuator exerts +8 N while a slider moves from x=0 to x=3 m, then exerts +2 N while it moves back to x=0. Forces are along x and constant within each leg.

1. Find work in each leg and the cycle total.

2. What work would result if +8 N acted on both legs?

3. Can zero net displacement alone establish zero work for the original schedule?

Question 4 - Solution

A force schedule matters even when displacement cancels

(a) Pair each force with its own displacement.

The governing relation is $$\boxed{W_{\mathrm{cycle}}=F_{\mathrm{out}}\Delta x_{\mathrm{out}}+F_{\mathrm{back}}\Delta x_{\mathrm{back}}}$$

$$W_1=8(3)=24\ \mathrm J,\qquad W_2=2(-3)=-6\ \mathrm J;$$ total is 18 J.

(b) Apply one unchanged force.

With +8 N both ways, total work is $$24-24=0\ \mathrm J.$$

(c) Keep the history of the interaction.

No. The original force is not constant across the complete trip. Different force-displacement pairings give different work even when the particle returns to its starting point.