Question

Given that a force $\overrightarrow{F}$ acts on a body for time $t$, and displaces the body by $\overrightarrow{d}$. In which of the following cases, the speed of the body must not increase?

• A. $|\overrightarrow{F}|>|\overrightarrow{d}|$
• B. $|\overrightarrow{F}|<|\overrightarrow{d}|$
• C. $\overrightarrow{F}=\overrightarrow{d}$
• D. $\overrightarrow{F}\perp \overrightarrow{d}$

Answer 1

Answer: (D)

$\overrightarrow{F}\perp \overrightarrow{d}$

Work done by a force $W=Fd\cos \theta$

where $F$ is the force acting on the body, $d$ is the displacement of the body and $\theta$ is the angle between $F$ and $d$.

If force is perpendicular to the displacement i.e. $\overrightarrow{F}$ $\perp \overrightarrow{S}$, then we get $\theta ={90}^o$

We know $\cos {90}^o=0$

$⟹$ $W=0$

Now from work energy theorem, change in kinetic energy of body $\Delta K.E=W$

$⟹$ $\Delta K.E=0$ which implies that velocity remains the same.

Hence velocity will remain same when force is perpendicular to the direction of displacement

i.e. $\hat{F}\perp \hat{d}$.