Question

Suppose two children push horizontally, but in exactly opposite directions, on a third child in a wagon. The first child exerts a force of 75.0 N, the second a force of 90.0 N, friction is 12.0 N, and the mass of the third child plus wagon is 23.0 kg.

(a) What is the system of interest if the acceleration of the child in the wagon is to be calculated?

(b) Draw a free-body diagram, including all forces acting on the system.

(c) Calculate the acceleration.

(d) What would the acceleration be if friction were 15.0 N?

Short answer

(a) The system here is the child inside the wagon along with the wagon.

(b) The free-body diagram is drawn as shown below.

(c) The value of acceleration is 0.13 m/s².

(d) The value of acceleration will be zero.

Step-by-step solution

(a) Determine the system of interest

For calculating the acceleration of the child inside the wagon, the system would be the child inside the wagon along with the wagon.

Given data

• Force exerted by the first child of 75.0 N.
• the second child exerts a force of 90.0 N.
• Friction is 12.0 N.
• the mass of the third child plus wagon is 23.0 kg.

(b) Draw the free body diagram of the system considering the forces acting on it

Draw the free-body diagram of the system along with the forces acting on it as shown below,

Here, W is the weight of the wagon plus the child, F₁ is the force exerted by the first child, F₂ is the force applied by the second child, f is the friction force, N is the normal reaction, m is the mass of wagon plus the third child.

(c) Calculate the acceleration

Apply Newton’s second law of motion.

$$\begin{gathered} F_{net}=ma \\ {F}_2-F_1-f=ma................\left(i\right) \end{gathered}$$

Here, F_net is the net force, and a is the acceleration.

Substitute 75Nfor F₁, 90N for F₂, 12N for f, and 23kg for m in equation (i), and we get,

$$\begin{gathered} 90\hspace{0.17em}N-75N-12N=23kg\times a \\ 3N=23Kg\times a \\ a=\frac{3kg.m/s^2}{23kg} \\ a=0.13m/s^{} \end{gathered}$$

Hence, the value of the acceleration is 0.13 m/s².

(d) Calculate the acceleration when the force of friction is 15N

Substitute 75Nfor F₁, 90N for F₂, 15N for f, and 23kg for m in equation (i), and we get,

$$\begin{gathered} 90N-75N-15N=23\mathrm{kg}\times a \\ 0=23\mathrm{kg}\times a \\ a=0 \end{gathered}$$

Hence, the acceleration will be 0.