Question

A force is inclined at ${60}^{\circ }$ to the horizontal produces a displacement of $\overrightarrow{s}=(\hat{i}+\hat{j})\text{m}$. If the component of force in horizontal direction is $50\text{N}$, the net work done by the force on the body is

• A. $-37\text{J}$
• B. $37\text{J}$
• C. $-137\text{J}$
• D. $137\text{J}$

Answer 1

The correct option is D $137\text{J}$

Given, Horizontal component of force $\left(F_x\right)=50\text{N}$
Displacement of the body $\left(\overrightarrow{s}\right)=(\hat{i}+\hat{j})\text{m}$

Since, $F_x=F\cos \theta =50\text{N}$ and $\theta ={60}^{\circ }$
$F=\frac{50}{\cos {60}^{\circ }}=100\text{N}$

Vertical component of force $\left(F_y\right)=F\sin \theta =100\sin {60}^{\circ }=50\sqrt{3}\text{N}$
So, $\overrightarrow{F}=(F_x\hat{i}+F_y\hat{j})=(50\hat{i}+50\sqrt{3}\hat{j})$
Net work done $\left(W\right)=\overrightarrow{F}.\overrightarrow{s}$
$W=(50\hat{i}+50\sqrt{3}\hat{j}).(\hat{i}+\hat{j})$
$W=50+50\sqrt{3}$, as $\sqrt{3}\approx 1.73$
$W=137\text{J}$

Hence option D is the correct answer