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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Laws of Motion
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Suppose A box has a mass of $100\; kg$, and you have to move it up a ramp. The ramp has an angle of $23^{\circ}$. What is the force needed to get the box moving up the ramp if the coefficient of static friction is $0.20$?

$\begin{array}{1 1} 562\;N \\ 36\;N\\24\;N \\62\;N\end{array} $

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Answer : $ 562\;N$
Force required = $F_g+F_s =m.g \sin \theta+\mu_s.m. g \;\cos \theta$
The equation to use is $m.g \sin \theta + \mu_s .m .g \cos \theta$
Substituting $m = 100\;kg, \theta = 23^{\circ}$ and $\mu_s = 0.2$, we get $F= 382 +180=562\;N$


answered Aug 28, 2014 by meena.p
edited Sep 11, 2014 by balaji.thirumalai

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