Question

Problems 68 and 69 show a free-body diagram. For each:

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!

Short answer

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

Part (b): The velocity of the car is $17.4m/s.$

Step-by-step solution

Part (a) Step 1: Given 

The free body diagram:

Part (a) Step 1: Writing realistic problem.

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

The suitable realistic problem is as follows

A toy car of weigh$9.8\mathrm{N}$is moving in a forward direction by the force of $20N$, the coefficient of friction between the toy car and the earth surface is$0.5$, determine the velocity of the car after $10m$if the car starts from rest. Part (b) Step 3: Given.

Part (b) Step 3 Given.

Weight of the car is $9.8\mathrm{N}.$

Forward force, $F=20\mathrm{N}$

Friction force,$F_{\text{friction}}=4.9\mathrm{N}.$

Part (b) Step 4: Formula used.

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:

$v^2=u^2+2as.$

Here,

$v$is the velocity.

$u$is the initial velocity.

$a$is the acceleration.

$s$is the distance.

Part (b) Step 5: Calculation.

The mass of the car is calculated as:

$\begin{array}{c}W=mg\\ 9.81=m\times 9.81\\ m=1\mathrm{kg}.\end{array}$

The acceleration of the car is calculated as

role="math" localid="1647781021275" $\begin{array}{c}{F}_{\text{net}}=ma\\ (20-4.9)=1\times a\end{array}$

$15.1\mathrm{m}/{\mathrm{s}}^2$

The Velocity of the car after $10m$is calculatedas:

role="math" localid="1647781193449" $\begin{array}{c}{v}^2=u^2+2aS\\ =0+2\times 15.1\times 10\\ v=17.4\mathrm{m}/\mathrm{s}.\end{array}$

Conclusion.

Part (a): A toy car of weight $9.8N$is moving in a forward direction by the force of $20N$, the coefficient of friction between the toy car and the earth surface is $0.5$, the velocity of the car after $10$m if the car starts from the rest.

Part (b): The velocity of the car is $17.4\mathrm{m}/\mathrm{s}.$