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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Work, Power and Energy
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A man cleaning a floor pulls a vacuum cleaner with a force of magnitude F = 50.0 N at an angle of $30.0^{\large\circ}$ with the horizontal. Calculate the work done by the force on the vacuum cleaner as the vacuum cleaner is displaced 3 m to the right.

$\begin{array}{1 1}130J\\150J\\120J\\100J\end{array} $

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Answer : 130J
The work done by the force is defined to be the product of component of the force in the direction of the displacement and the magnitude of this displacement. Thus W = $(F \cos \theta )d = F.d$
Using the definition of work
W = $50 \times \cos 30 \times 3$
$\;\;\;=130N.m$
$\;\;\;=130J$
answered Aug 28, 2014 by sreemathi.v
 

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