Work — Question 3

Question 3

A reversing resistance can do work around a closed trip

A block slides 4 m right and then 4 m left on a fixed floor. Kinetic friction has constant magnitude 3 N and always opposes sliding. Ignore the short turnaround when computing work.

1. Find friction work on the outward leg.

2. Find friction work on the return and for the whole trip.

3. Why is multiplying one friction vector by the zero net displacement invalid?

Question 3 - Solution

A reversing resistance can do work around a closed trip

(a) Use the outward force.

The governing relation is $$\boxed{W_f=\int\vec f\cdot d\vec r=-f\int ds=-fL}$$ Outward friction is $-3\ \mathrm N$ and displacement is +4 m, giving $$W_1=-12\ \mathrm J.$$

(b) Change both signs on the return.

Return friction is +3 N and displacement $-4\ \mathrm m$, giving $$W_2=-12\ \mathrm J.$$ Total work is $-24\ \mathrm J$.

(c) Test the constant-force assumption.

The friction vector changed direction at reversal. It cannot be factored out of the segment sum. Work here depends on total sliding distance: $$W_f=-3(8)=-24\ \mathrm J$$ despite zero net displacement.