STEP 1
Establish relationship between force and acceleration.
From the 1st statement, there is a direct variation between the force and the acceleration. This is put mathematically as
[tex]F\text{ }\alpha\text{ a}[/tex]We introduce a constant, m
[tex]\begin{gathered} F\text{ = ma where} \\ m\text{ is the proportionality constant creating the relationship thus } \\ Nperm/s^2\text{ as its unit} \end{gathered}[/tex]STEP 2
Derive value for the constant, m
[tex]\begin{gathered} F\text{ = ma} \\ \text{Dividing both sides by a, we have} \\ \frac{F}{a}=m \\ where\text{ }f=56N,m=8m/s^2 \\ m=\frac{56}{8}=7Nperm/s^2=7Ns^2\text{ / m} \end{gathered}[/tex]STEP 3
Apply this value of m to solve related equations.
[tex]\begin{gathered} \text{where a = 3m/s}^2\text{ and m = }7Ns^2\text{ /m} \\ \text{and F = ma = 3 x 7 = 21}N \end{gathered}[/tex]Thus, the force when acceleration becomes 3 m/sq seconds is 21 N