$\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$

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