Question

1. 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 kgs. 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?

Explanation:

Step1: Identify system of interest

The system of interest is the third - child plus the wagon as we want to find its acceleration.

Step2: Calculate net force for part c

The two applied forces are in opposite directions. Let the force of the second child be positive. Net force $F_{net}=F_2 - F_1 - f$. Here, $F_1 = 75.0\ N$, $F_2 = 90.0\ N$ and $f = 12.0\ N$. So $F_{net}=90.0 - 75.0-12.0=3.0\ N$. According to Newton's second law $F_{net}=ma$, where $m = 23.0\ kg$. Then $a=\frac{F_{net}}{m}$.

Step3: Calculate acceleration for part c

$a=\frac{3.0}{23.0}\approx0.13\ m/s^{2}$.

Step4: Calculate net force for part d

If $f = 15.0\ N$, then $F_{net}=F_2 - F_1 - f=90.0 - 75.0-15.0 = 0\ N$. According to $F_{net}=ma$, when $F_{net}=0$, $a = 0\ m/s^{2}$.

Answer:

a) The third - child plus the wagon.
b) (A free - body diagram cannot be drawn in this text - based format. But it would have a right - ward force of 90.0 N, a left - ward force of 75.0 N, a left - ward friction force, and vertical forces of weight downwards and normal force upwards which balance each other as there is no vertical acceleration.)
c) $0.13\ m/s^{2}$
d) $0\ m/s^{2}$