Question

A Force $F$ making an angle ${60}^{\circ }$ with the horizontal, acts on an object and displaces it by $0.4\text{m}$ along the horizontal direction. If the work done by the force is $1\text{Joule}$, then the force $F$ is

• A. $1.5\text{N}$
• B. $2.5\text{N}$
• C. $3.5\text{N}$
• D. $5.0\text{N}$

Answer 1

The correct option is D $5.0\text{N}$


The horizintal component of the applied force is $Fcos\theta$

Work done by force on the body $=1\text{J}$

$W=\overrightarrow{F}.\overrightarrow{S}$

$\Rightarrow W=|F||S|cos\theta$

$1=|F|\left(0.4\right)cos{60}^{\circ }$

$F=5\text{N}$

Why this Question? Note: Only that component of force contributes in the calculation of work done which is along the displacement. The component of force perpendicular to the displacement contributes zero work.