First we have draw the free body diagram

The angle between horizontal and inclined plane is \(\displaystyle\theta={18}^{\circ}\)

a) From the diagram we can write

\(\displaystyle{m}{g}{\sin{\theta}}+{f}_{{s}}={m}{a}\)

\(\displaystyle{m}{g}{\sin{\theta}}+\mu_{{s}}{m}{g}{\cos{\theta}}={m}{a}\)

Therefore the magnitude od acceleration

\(\displaystyle{a}={g{{\left({\sin{\theta}}+\mu_{{s}}{m}{g}{\cos{\theta}}={m}{a}\right.}}}\)

b) For this in the diagram the direction offriction reversed.

Therefore we can write

\(\displaystyle{m}{g}{\sin{\theta}}-\mu_{{s}}{m}{g}{\cos{\theta}}={m}{a}\)

Therefore the magnitude of acceleration

\(\displaystyle{a}={g{{\left({\sin{\theta}}-\mu_{{s}}{\cos{\theta}}={m}{a}\right.}}}\)

substitute the given values.

The angle between horizontal and inclined plane is \(\displaystyle\theta={18}^{\circ}\)

a) From the diagram we can write

\(\displaystyle{m}{g}{\sin{\theta}}+{f}_{{s}}={m}{a}\)

\(\displaystyle{m}{g}{\sin{\theta}}+\mu_{{s}}{m}{g}{\cos{\theta}}={m}{a}\)

Therefore the magnitude od acceleration

\(\displaystyle{a}={g{{\left({\sin{\theta}}+\mu_{{s}}{m}{g}{\cos{\theta}}={m}{a}\right.}}}\)

b) For this in the diagram the direction offriction reversed.

Therefore we can write

\(\displaystyle{m}{g}{\sin{\theta}}-\mu_{{s}}{m}{g}{\cos{\theta}}={m}{a}\)

Therefore the magnitude of acceleration

\(\displaystyle{a}={g{{\left({\sin{\theta}}-\mu_{{s}}{\cos{\theta}}={m}{a}\right.}}}\)

substitute the given values.