Answer:
Approximately [tex]0.06\; {\rm m \cdot s^{-2}}[/tex], assuming that the force from the pool cue is horizontal.
Explanation:
As the billiard ball moves forward, the pool table will exert friction on the ball. In this question, it is given that the magnitude of this friction is [tex]7\; {\rm N}[/tex]. If the ball is moving forward, this friction will point backward.
The cue exerts a forward force of magnitude [tex]22\; {\rm N}[/tex] on the billiard ball. The [tex]22\; {\rm N}[/tex] forward force from the cue and the [tex]7\; {\rm N}[/tex] backward force from the table interact with each other. The resultant force on the ball will be [tex](22\; {\rm N} - 7\; {\rm N}) = 15\; {\rm N}[/tex], forward.
When an object of mass [tex]m[/tex] experiences a net force of [tex]F[/tex], the acceleration [tex]a[/tex] of that object will be [tex]a = (F / m)[/tex].
Apply unit conversion and ensure that the mass of the billiard ball is in standard units (grams): [tex]m = 0.25\; {\rm kg} = 250\; {\rm g}[/tex].
In this question, the net force on this billiard ball is [tex]F = 15\; {\rm N}[/tex]. With a mass of [tex]m = 250\; {\rm g}[/tex], the acceleration [tex]a[/tex] of this billiard ball will be:
[tex]\begin{aligned}a &= \frac{F}{m} \\ &= \frac{15\; {\rm N}}{250\; {\rm g}} = 0.06\; {\rm m\cdot s^{-2}}\end{aligned}[/tex].
