Given
Mass is m=2.61 kg
Angle is
[tex]\theta=15^o[/tex]Coefficient of friction is,
[tex]\mu_k=0.22[/tex]To find
The acceleration of the block
Explanation
The free body diagram is
Thus at equilibrium,
[tex]\begin{gathered} ma=mgsin15-f \\ ma=mgsin15-\mu_kmgcos15 \\ \Rightarrow a=9.8sin15-0.22\times9.8cos15 \\ \Rightarrow a=0.45\text{ m/s}^2 \end{gathered}[/tex]Conclusion
The acceleration is
[tex]0.45\text{ m/s}^2[/tex]
