The given problem can be exemplified in the following diagram:
Since there is no friction or any other external force, the only force acting in the direction of the movement is the component of the weight of the block, therefore, applying Newton's second law:
[tex]\Sigma F=ma[/tex]
Replacing the values:
[tex]mg\sin 40=ma[/tex]
We may cancel out the mass:
[tex]g\sin 40=a[/tex]
Using the gravity constant as 9.8 meters per square second:
[tex]9.8\frac{m}{s^2}\sin 40=a[/tex]
Solving the operations:
[tex]6.3\frac{m}{s^2}=a[/tex]
Therefore, the acceleration is 6.3 meters per square second.
