Question

A pool cue strikes a 0.25 kg billiard ball with a force of 22 N. If the felt on the pool table supplies

a force of 7 N what is the acceleration of the billiard ball.

Answer 1

Approximately
`0.06\; {\rm m \cdot s^(-2)}`, assuming that the force from the pool cue is horizontal.

Step-by-step 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
`7\; {\rm N}`. If the ball is moving forward, this friction will point backward.

The cue exerts a forward force of magnitude
`22\; {\rm N}` on the billiard ball. The
`22\; {\rm N}` forward force from the cue and the
`7\; {\rm N}` backward force from the table interact with each other. The resultant force on the ball will be
`(22\; {\rm N} - 7\; {\rm N}) = 15\; {\rm N}`, forward.

When an object of mass
`m` experiences a net force of
`F`, the acceleration
`a` of that object will be
`a = (F / m)`.

Apply unit conversion and ensure that the mass of the billiard ball is in standard units (grams):
`m = 0.25\; {\rm kg} = 250\; {\rm g}`.

In this question, the net force on this billiard ball is
`F = 15\; {\rm N}`. With a mass of
`m = 250\; {\rm g}`, the acceleration
`a` of this billiard ball will be:

`\begin{aligned}a &= (F)/(m) \\ &= \frac{15\; {\rm N}}{250\; {\rm g}} = 0.06\; {\rm m\cdot s^(-2)}\end{aligned}`.