Work Done by the Gravitational Force — Question 1

Question 1

Gravity counts signed height changes along a multistage route

Take a uniform gravitational field with $g=10$ m/s$^2$, upward y positive, and neglect air resistance. A 2 kg parcel moves from y=1 m up to y=5 m, then down to y=3 m. Other agents prescribe the motion.

1. Find gravitational work on each segment.

2. Find total gravitational work and compare with a direct move from y=1 to y=3 m.

3. Why is $-mg$ times the total vertical distance not the correct total work?

Question 1 - Solution

Gravity counts signed height changes along a multistage route

(a) Use the signed displacement in each segment.

The governing relation is $$\boxed{W_g=\int_{y_i}^{y_f}(-mg)\,dy=-mg(y_f-y_i)}$$

$$W_{g,1}=-20(5-1)=-80\ \mathrm J,\qquad W_{g,2}=-20(3-5)=+40\ \mathrm J.$$

(b) Add and telescope the heights.

Total is $-40\ \mathrm J$, equal to $-20(3-1)$. In a uniform field, $$W_g=-mg(y_f-y_i)$$ for any route.

(c) Keep upward and downward contributions distinct.

Total vertical distance is 6 m, but gravity does negative work during ascent and positive work during descent. Assigning the same negative sign to both would wrongly give $-120\ \mathrm J$.