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 exerts 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?
a) The third child in the wagon
b) The first child pushing
c) The second child pushing
d) The combined system of all three children and the wagon

(B) Calculate the acceleration.
a) 3.26 m/s²
b) 3.78 m/s²
c) 4.05 m/s²
d) 3.95 m/s²

Answer 1

The system of interest for calculating the child's acceleration is the combined system of the wagon and the third child. The combined system experiences a net force resulting in acceleration when taking into account both opposing forces from the children and the force of friction. Using Newton's second law, the acceleration is found to be 0.130 m/s² in the direction of the second child's push.

Step-by-step explanation:

When calculating the acceleration of the child in the wagon, we consider the combined system of the third child plus the wagon. This system of interest includes all the forces acting on it, which are the external forces exerted by the two children and the force of friction.

To calculate the acceleration (a), we first determine the net force (Fnet) by subtracting the opposing forces and considering the force of friction. The first child exerts a force of 75.0 N and the second exerts 90.0 N, giving us a net force of 90.0 N - 75.0 N - 12.0 N (friction) = 3.0 N in the direction of the second child's push.

Using Newton's second law, Fnet = m × a, we can solve for acceleration: a = Fnet / m = 3.0 N / 23.0 kg = 0.130 m/s². Therefore, the acceleration of the system is 0.130 m/s² in the direction of the second child's push.