Question

Michael applies a 8 N rightward force to a 0.75 kg book to accelerate it across a table top. The coefficient of friction between the book and the tabletop is 0.610. Determine the acceleration of the book.

Answer 1

The Newton's second law says

$$a=\frac{F_{\rm net}}{m}=\frac{F-\mu mg}{m}$$

$$a=\frac{8-0.610*9.8*0.75}{0.75}=4.7\:\rm m/s^2$$