$\begin{array}{1 1} 1.146\;m/s^2 \\4.146 \;m/s^2\\8.14\;m/s^2 \\1.6\;m/s^2\end{array} $

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Answer : $1.146\;m/s^2$

Draw a free body diagram

At this point , We are as to the overall direction of motion , therefore we cannot correctly label $F_f$ on the diagram .

Step 2:

$F_{appx} =15 \cos 32=12.7207\;N$

$F_{appy} =15 \sin 32=7.9488\;N$

$F_{gx}=mg \sin \theta =(6.0)(9.80) \sin 32$

$\qquad= 31.1592 \;N$

$F_{appy}=15 \sin 32 =7.9488\;N$

$F_{gy} =mg \cos \theta =(6.0) (9.80) \cos 32$

$\qquad=49.8652\;N$

Step 3:

$F_N= F_{gy}+F_{appy }$

$\qquad= 49.8652+7.9488$

$\qquad= 57.8139\;N$

Therefore : $ F_f= \mu F _N$

$\qquad= (0.26) (57.8139\;N)$

$\qquad=11.5627 \;N$

Step 4:

We will call $F_f$ negative as the overall motion is down the ramp (in the direction of the force of gravity Since $F_{gx} > F_{app}$) , therefore the force of friction is directed in the opposite direction of the motion (up the ramp)

$F_{net}=F_{gx}-F_{appx} - F_{f}$

$\qquad=31.1592 \;N -12.7207 \;N -11.5627\;N =6.8758\;N$

$a=F_{net}/m =6.8758/6.0 =1.146\;m/s^2$

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