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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Laws of Motion
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A $12-kg$ box is released from the top of an inclined that is $5.0\;m$ long and makes an angle of $40^{\circ}$ to the horizontal. A $60-N$ friction force impedes the motion of the box. How long will it take to reach the bottom of the incline ?

$\begin{array}{1 1}2.76\;s\\36\;s\\24\;s \\62\;s\end{array} $

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Answer : $2.76\;s$
We show the three forces acting on the block : the friction force $f=60\;N;$
the normal force N, which is perpendicular to the incline; and the weight of the block, $w=mg =(12\;kg)(9.8 m/s^2) =118\;N$
We choose the x-axis along the incline with downward as positive .
Using $\sum F_x =ma_x,$ we have $w \sin 40^{\circ}-f=ma_x$ or $(118 \;N) (0.642) -(60\;N)=(12\;kg)a_x$.
Solving we have $a_x = 1.31 m/s^2 $
To find the time to reach the bottom of the incline , starting from rest , we use $x= v_{0x}t +\large\frac{1}{2}$$a_x t^2$, with $v_{ax} =0$ and $x= 5.0 \;m$
Solving we get $t=(7.63 \;s^2)^{1/2}$
$ \qquad= 2.76\;s$
answered Aug 28, 2014 by meena.p
edited Sep 11, 2014 by meena.p
 

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