Question
Figure shows three forces applied to a trunk that moves leftward by $3\, m$ over a smooth floor. The force magnitudes are $F_1 = 5\,N, F_2 = 9\,N$, and $F_3 = 3\,N$. The net work done on the trunk by the three forces $......... \mathrm{J}$
✓
Answer
$\overrightarrow F = - 5\hat i + 9\cos {60^ \circ }\hat i + 9\sin {60^ \circ }\hat j - 3\hat j$
$= - 5\hat i + \frac{9}{2}\hat i + \frac{{9\sqrt 3 }}{2}\hat j - 3\hat j$
$= - \frac{{\hat i}}{2} + \left( {\frac{{9\sqrt 3 }}{2} - 3} \right)\hat j$
$\overrightarrow s = - 3\hat i.$
$W = \overrightarrow F \overrightarrow s = \left[ { - \frac{{\hat i}}{2} + \left( {\frac{{9\sqrt 3 }}{2} - 3} \right)\hat j} \right].\left( { - 3\hat i} \right)$
$= 1.5\,J.$
$= - 5\hat i + \frac{9}{2}\hat i + \frac{{9\sqrt 3 }}{2}\hat j - 3\hat j$
$= - \frac{{\hat i}}{2} + \left( {\frac{{9\sqrt 3 }}{2} - 3} \right)\hat j$
$\overrightarrow s = - 3\hat i.$
$W = \overrightarrow F \overrightarrow s = \left[ { - \frac{{\hat i}}{2} + \left( {\frac{{9\sqrt 3 }}{2} - 3} \right)\hat j} \right].\left( { - 3\hat i} \right)$
$= 1.5\,J.$
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