Given free body diagram :

a. Write a realistic dynamics problem for which this is the correct free-body diagram. Your problem should ask a question that can be answered with a value of position or velocity (such as “How far?” or “How fast?”), and should give sufficient information to allow a solution.

b. Solve your problem!

The free body diagram :

In the figure, there is an downward force of $9.8\mathrm{N}$which signifies the weight of the car and there is an upward force which represent the reaction force, the forward force which tend to move the toy car in forward direction and a back ward force which oppose the motion of the car due to friction.

The suitable realistic problem according to the free body diagram is as follows ;-

A toy car of weight $9.8\mathrm{N}$is moving in a forward direction by the force of $20\mathrm{N}$, the coefficient of friction between the toy car and the earth surface is $0.5$, determine the velocity of the car after $10\mathrm{m}$if the car start from rest.

The required problem is : A toy car of weight $9.8\mathrm{N}$is moving in a forward direction by the force of $20\mathrm{N}$, the coefficient of friction between the toy car and the earth surface is $0.5,$determine the velocity of the car after $10\mathrm{m}$if the car start from rest.

Weight of the car = $9.8\mathrm{N}$, forward force, $F=20\mathrm{N}$, friction force, $F_{\text{friction}}=4.9\mathrm{N}.$

The mass of the truck is calculated as : $m=\frac{W}{g}.$

Here,

W is weight

g is acceleration due to gravity

The acceleration of the truck is calculated as : $F=ma$

Here,

a is the acceleration.

m is the mass of the truck.

The velocity of the truck is calculated as : role="math" $v^2=u^2+2as.$

Here,

v is the velocity, u is the initial velocity, a is the acceleration, s is the distance.

The mass of the car is calculated as :-

$$\begin{gathered} W=mg \\ 9.81=m\times 9.81 \\ m=1\mathrm{kg} \end{gathered}$$

The acceleration of the car is calculated as :-

$$\begin{gathered} F_{\text{net}}=ma(20-4.9) \\ =1\times a15.1\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

The velocity of the car after $10\mathrm{m}$is calculated as

$$\begin{gathered} v^2=u^2+2aS \\ =0+2\times 15.1m/s^2\times 10s \\ v=17.4\mathrm{m}/\mathrm{s} \end{gathered}$$

The velocity of the car is$17.4\mathrm{m}/\mathrm{s}.$