Question

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

• A. $1.50\text{J}$
• B. $2.40\text{J}$
• C. $3.00\text{J}$
• D. $6.00\text{J}$

Answer 1

The correct option is A $1.50\text{J}$

Consider unit vector along right as $\hat{i}$ and unit vector upwards as $\hat{j}$
Net Force acting on the body is$\overrightarrow{F}=\overrightarrow{F_1}+\overrightarrow{F_2}+\overrightarrow{F_3}\overrightarrow{F}=-5\hat{i}+9\text{cos}{60}^{\circ }\hat{i}+9\text{sin}{60}^{\circ }\hat{j}-3\hat{j}$
$\overrightarrow{F}=-5\hat{i}+\frac{9}{2}\hat{i}+\frac{9\sqrt{3}}{2}\hat{j}-3\hat{j}$
$\overrightarrow{F}=-\frac{\hat{i}}{2}+(\frac{9\sqrt{3}}{2}-3)\hat{j}$
Given displacement is
$\overrightarrow{s}=-3\hat{i}$.
Work done by net force
$W=\overrightarrow{F}.\overrightarrow{s}$
$W=[-\frac{\hat{i}}{2}+(\frac{9\sqrt{3}}{2}-3)\hat{j}].(-3\hat{i})$
$W=1.5\text{J}$.