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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Work, Power and Energy
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A horizontal force F pulls a 20kg box at a constant velocity along a horizontal floor.If the coefficient of friction between the box and the floor is 0.25,find the work done by force F in moving the box through a distance of 2m.

$\begin{array}{1 1}98J\\78J\\108J\\88J\end{array} $

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Answer : 98J
Since the box is moved at a constant velocity,the applied force F just overcomes the frictional force f,i.e,
$F=f=\mu mg$
$\therefore$ Work done $W =FS\cos\theta=\mu mgS\cos 0^{\large\circ}$
$\Rightarrow 0.25\times 20\times 9.8\times 2 =98J$
answered Aug 28, 2014 by sreemathi.v
 

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