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A constant force of 45 N directed at angle θ to the horizontal pulls a crate of weight 100 N from one end of a room to another a distance of
Question
A constant force of 45 N directed at angle θ to the horizontal pulls a crate of weight 100 N from one end of a room to another a distance of 4 m. Given that the vertical component of the pulling force is 12 N, calculate the work done by the force in moving the crate.
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Physics
3 years
2021-07-31T15:02:14+00:00
2021-07-31T15:02:14+00:00 1 Answers
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Answer:
Explanation:
information we know:
Total force:![Rendered by QuickLaTeX.com F=45N](https://documen.tv/wp-content/ql-cache/quicklatex.com-898b615259438f67f5df411efd8cf47c_l3.png)
Weight:![Rendered by QuickLaTeX.com w=100N](https://documen.tv/wp-content/ql-cache/quicklatex.com-ef1e717125bb346d8d028863757a2d75_l3.png)
distance:![Rendered by QuickLaTeX.com 4m](https://documen.tv/wp-content/ql-cache/quicklatex.com-cf2c188f034fcb91518330ab6a1759f4_l3.png)
vertical component of the force:![Rendered by QuickLaTeX.com F_{y}=12N](https://documen.tv/wp-content/ql-cache/quicklatex.com-92fe6e6b32893569d51d8e47fd83f9d4_l3.png)
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In this case we need the formulas to calculate the components of the force (because to calculate the work we need the horizontal component of the force).
horizontal component:![Rendered by QuickLaTeX.com F_{x}=Fcos\theta](https://documen.tv/wp-content/ql-cache/quicklatex.com-2ad5052959b41d081aa1cd6c33ef03a0_l3.png)
vertical component:![Rendered by QuickLaTeX.com F_{y}=Fsen\theta](https://documen.tv/wp-content/ql-cache/quicklatex.com-346279555adfd9a036b1344d8abeee51_l3.png)
but from the given information we know that![Rendered by QuickLaTeX.com F_{y}=12N](https://documen.tv/wp-content/ql-cache/quicklatex.com-92fe6e6b32893569d51d8e47fd83f9d4_l3.png)
so, equation these two
and ![Rendered by QuickLaTeX.com F_{y}=12N](https://documen.tv/wp-content/ql-cache/quicklatex.com-92fe6e6b32893569d51d8e47fd83f9d4_l3.png)
and we know the force
, thus:
now we clear for![Rendered by QuickLaTeX.com \theta](https://documen.tv/wp-content/ql-cache/quicklatex.com-356a08e839ab6974a16448e16e56745d_l3.png)
the angle to the horizontal is 15.466°, with this information we can calculate the horizontal component of the force:
whith this horizontal component we calculate the work to move the crate a distance of 4 m:
the work done is W=173.48J