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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Work, Power and Energy
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A horizontal force F pulls a 20-kg carton across the floor at constant speed.If the coefficient of sliding friction between carton and floor is 0.60,how much work does F do in moving the carton 3.0m?

$\begin{array}{1 1}353J\\253J\\153J\\453J\end{array} $

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Answer : 353J
Because horizontal speed is constant,the carton is in horizontal equilibrium.
$F = f=\mu F_N$
Normal force is the weight,$20(9.8) = 196N$
Therefore $W =Fx=0.60(196)(3.0)=353J$
answered Aug 28, 2014 by sreemathi.v
 

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