# Two children pushing on a third child in a wagon| Newton’s Second Law of Motion: Concept of a System| Dynamics| College Physics| Problem 4.9

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

#### SOLUTION:

PART A

If the acceleration of the child in the wagon is to be calculated, the system of interest is the wagon with the child in it.

PART B

Considering all the forces acting on the wagon and child with mass m. there will be 5 forces. The horizontal forces are the two forces coming from the two other children, denoted as $F_l\:$ and $F_r$. We also have friction going against the direction of motion.

The vertical forces are the weight of the wagon and the child and the normal force. These forces are shown in the free-body diagram shown.

PART C

From Newton’s second law of motion, we know that

$net\:F=ma$

Considering the horizontal direction to the right as positive, the forces involved are $F_r,\:F_l\:and\:f$. Therefore, we have

$F_r-F_l-f=ma$

Solving for acceleration, we have

$a=\frac{F_r-F_l-f}{m}$

$a=\frac{90.0\:N-75.0\:N-12.0\:N}{23.0\:kg}$

$a=0.1304\:m/s^2$

The direction of the motion is towards the direction of the child pushing at 90.0 N.

PART D

If the friction is 15 N, the acceleration would be

$a=\frac{90.0\:N-75.0\:N-15.0\:N}{23.0\:kg}$

$a=0\:m/s^2$

There would be no acceleration, meaning that the system is in equilibrium. The system is either not moving or it’s moving at a constant velocity.