Question

Why is work minimum in case of direction of body displaced is perpendicular to the direction force applied and not in the case when the direction of body displaced is opposite to direction of force applied?

Answer 1

Step 1: Given parameters

In case, the body is displaced perpendicular to the direction of the force.

The angle between force and displacement, $\theta =90°$

Step 2: Formula Used

Calculate the work done by using the formula given as,

$W=F·d$

Where $W$ is work done, $F$ is force, and $d$ is displacement

Step 3: Calculation

Work done is the dot product of force and displacement.

$W=F·d$

$W=Fd\cos \theta$

When the direction of displacement is perpendicular to the direction of force so, $\theta =90°$

$$\begin{gathered} W=Fd\cos 90° \\ W=Fd\left(0\right) \\ W=0 \end{gathered}$$

So, the work done is zero.

In case, the body is displaced opposite to the direction of the force.

The angle between force and displacement, $\theta =180°$

$W=F·d$

$W=Fd\cos \theta$

When the direction of displacement is opposite to the direction of force so, $\theta =180°$

$$\begin{gathered} W=Fd\cos 180° \\ W=Fd(-1) \\ W=-Fd \end{gathered}$$

Here the minus sign indicates the negative work done which means the force is opposite to the displacement of the object.

When the direction of displacement is perpendicular to the direction of force, the work done is zero, but when the displacement is in the opposite direction of force, the work done is negative.

Hence the minimum work will be done when the direction of displacement and force are perpendicular to each other.