Question

4. A 1.2 kg block of mass is placed on an inclined plane that has a slope of 30° with respect to the horizontal, and the block of mass is released. A friction force Fk = 2.0 N acts on the block as it slides down the incline. Find the acceleration of the block down the incline

Answer 1

The free body diagram of the problem has 3 forces:
1) The Normal perpendicular to inclined plane. n=m*g=1.2*10=12 N
2) The friction parallel to inclined plane f=2.0 N
3) The weight that has two components, one perpendicular and other parallel (n*sin(30))
The Newton equation parallel to inclined plane is:
\[-f + m \times g \sin (30) = m \times a\]
\[a=acceleration=-f \div m + g \times \sin (30)= -2 \div 1.2 + 5 = 3.33 meters/\sec^{2}\]