Question

Figure shows three forces applied to a block that moves leftward by $3\text{m}$ over a smooth floor. The magnitudes of force are $F_1=5\text{N},F_2=9\text{N}$ and $F_3=3\text{N}$. The net work done on the block by the three forces will be

• A. $1.5\text{J}$
• B. $3.6\text{J}$
• C. $3\text{J}$
• D. $4.5\text{J}$

Answer 1

The correct option is A $1.5\text{J}$


From the FBD of body and considering the rightwards as $+vex-$ direction and upwards as $+vey-$ direction ,
$\overrightarrow{F}=-5\hat{i}+9\cos {60}^{\circ }\hat{i}+9\sin {60}^{\circ }\hat{j}-3\hat{j}$
$=-5\hat{i}+\frac{9}{2}\hat{i}+\frac{9\sqrt{3}}{2}\hat{j}-3\hat{j}$
$=-5\hat{i}+\frac{9}{2}\hat{i}+\frac{9\sqrt{3}}{2}\hat{j}-3\hat{j}$
$\overrightarrow{F}=-\frac{1}{2}\hat{i}+(\frac{9\sqrt{3}}{2}-3)\hat{j}$
Displacement of the block $\overrightarrow{S}=-3\hat{i}\text{m}$
Work done by force,
$W=\overrightarrow{F}\cdot \overrightarrow{S}=[-\frac{1}{2}\hat{i}+(\frac{9\sqrt{3}}{2}-3)\hat{j}]\cdot (-3\hat{i})$
$\therefore W=1.5\text{J}$