Question

Write the expression of the work done against the gravitational force of mass m in moving mass m from r1 to r2

Answer 1

Answer & Explanation:

The gravitational force is represented by a vector directed vertically downward. The magnitude of the gravitational force on a body of mass $m$ is given by $mg$, where $g$ is the acceleration of gravity. If we assume that $\mathbf{u}_{y}$ is the unit vector pointing vertically upward, we can write the gravitational force in vector form as

$\mathbf{F}_g = - \, m g \mathbf{u}_{y} \, \, .$

The work done by the gravitational force between is given by

$W = \int_{\mathbf{r}_1}^{\mathbf{r}_2} \mathbf{F}_g \cdot \mathrm{d} \mathbf{r} \, \, .$

We can decompose the infinitesimal element as

$\mathrm{d} \mathbf{r} = \mathrm{d} x \, \mathbf{u}_x + \mathrm{d} y \, \mathbf{u}_y \, \, .$

where $\mathbf{u}_{x}$ is the unit vector in the horizontal, such that

$\mathbf{u}_{y} \cdot \mathbf{u}_{y} = 1 \quad \text{and} \quad \mathbf{u}_{y} \cdot \mathbf{u}_{x} = 0 \, \, .$

Now we proceed with the calculation,

$W = \int_{\mathbf{r}_1}^{\mathbf{r}_2} \left( - \, m g \mathbf{u}_{y} \right) \cdot \left( \mathrm{d} x \, \mathbf{u}_x + \mathrm{d} y \, \mathbf{u}_y \right) \\ \\= \int_{y_1}^{y_2} \left( - \, m g \right) \mathrm{d} y \right) \\ \\= - m g ( y_2 - y_1 )\, \, .$

In the above calculation we changed the limits of integration to include only the $y$ component of the vector.

We can see that as we raise a body of mass $m$ the work done by gravity is negative, since in this case $y_2 - y_1 > 0$. When the body falls, $y_2 - y_1 < 0$ and the work done by gravity is positive.