Work Done by the Gravitational Force — Question 7

Question 7

Gravity work follows the field direction in rotated coordinates

In a perpendicular but tilted laboratory coordinate system, uniform gravitational acceleration has components $(6,-8)$ m/s$^2$. A 2 kg particle is displaced by $(4,3)$ m. Other agents may prescribe the route.

1. Find gravity work for this displacement.

2. Explain the geometric meaning of the result.

3. Find gravity work for displacement $(0,-4)$ m in the same coordinates.

Question 7 - Solution

Gravity work follows the field direction in rotated coordinates

(a) Use the complete gravitational force vector.

The governing relation is $$\boxed{W_g=m\vec g\cdot\Delta\vec r}$$

$$\vec F_g=2(6,-8)=(12,-16)\ \mathrm N.$$ Thus $$W_g=12(4)-16(3)=0.$$

(b) Check perpendicularity to the field.

$$(6,-8)\cdot(4,3)=0,$$ so the displacement is perpendicular to true down. A change in the coordinate labeled y need not be a change in physical height when the axes are tilted.

(c) Project the second displacement.

$$W_g=12(0)+(-16)(-4)=64\ \mathrm J.$$ Its magnitude is less than $$mg(4)=80\ \mathrm J$$ because the displacement is not exactly along gravity.