$\begin{array}{1 1}-9.2\;m/s^2\\6.7\;m/s^2\\7.5 \;m/s^2\\10\;m/s^2\end{array} $

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Step 2:

The component of gravity , $F_g$

$F_{gx} =F_g \sin \theta$

$F_{gx}=mg \sin \theta$

$\qquad= (7.0)(9.80) \sin 35$

$\qquad= 39.3473\;N$

Step 3:

The force of friction , $F_f$

$F_f= \mu _k F_N$

$F_f = \mu _k mg \cos \theta$

$\qquad= (0.45) (7.0)(9.80) \cos 35$

$\qquad= 25.2872\;N$

Step 4:

In this case, both the forces are opposite to the direction of motion, and are therefore both negative.

$F_{net} =-F_{gx} -F_f$

$F_{net }= -39.3473 \;N- 25.2872=-64.6345$

$a= F_{net/m}=-64.6345/7.0$

$a= F_{net /m} =-64.6345/7.0 =-9.2 m/s^2$

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