Question

What is the significance of the negative sign in $$W=-mgh$$ ?

Answer 1

Hint:Work is the measure of transfer of energy when one object is displaced from one point to another by an external force. This external force should be fully or partially directed towards the direction of displacement.

Formulae used:
$W=\overrightarrow{F}\cdot \overrightarrow{d}$
where $W$ is the work done, $\overrightarrow{F}$ is the force vector doing the work and $\overrightarrow{d}$ is the displacement.
$\overrightarrow{F}=m\cdot \overrightarrow{a}$
where $\overrightarrow{F}$ is the force vector, $m$ is the mass of the body on which force is applied and $\overrightarrow{a}$ is the acceleration vector.

Complete step by step answer:
Work done is the dot product of two vectors, namely Force and Displacement. Hence, work done is a scalar quantity. Therefore it is devoid of any direction.From work done equation, we have
$$\begin{gathered} \Rightarrow W=\overrightarrow{F}\cdot \overrightarrow{d} \\ \Rightarrow W=Fd\cos \theta \end{gathered}$$
Where $\theta$ is the angle between force and displacement vector.

Substituting $\overrightarrow{F}$ as $\overrightarrow{F}=m\cdot \overrightarrow{a}$ , we get
$\Rightarrow W=mad\cos \theta$
Acceleration here can be taken as $g$ (acceleration due to gravity) and displacement as height $h$ from the surface of the earth. Also, now $\theta$ becomes the angle between displacement and acceleration vector.Therefore, the equation becomes
$$W=mgh\cos \theta$$

When we consider the surface of the earth as origin and $g$ down towards the earth as positive, then any displacement from the surface of the earth to the sky will result in the angle between $g$ and $h$ to become ${180}^{\circ }$.Now, the value of $\cos {180}^{\circ }=-1$. Hence, the equation becomes
$$W=-mgh$$
Where the negative sign denotes that work is done against gravity.

Note:This result holds true only if the body being dropped, or raised is close to the Earth’s surface. This is because as the distance from the surface of the earth increases, the magnitude of gravitational force changes. Hence the value of $g$ starts changing.