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.

The net force acting on the third child plus wagon is approximately:
a)3.0 , {N}
b)15.0 , {N}
c)24.0 , {N}
d)165.0 , {N}

Answer 1

The net force acting on the child plus wagon system is 153.0 N, and the acceleration of the child in the wagon is 6.65 m/s².

Step-by-step explanation:

To calculate the acceleration of the child in the wagon, we first need to determine the net force acting on the child plus wagon system. This can be found by subtracting the frictional force from the sum of the forces exerted by the two children. In this case, the first child exerts a force of 75.0 N, the second child exerts a force of 90.0 N, and the frictional force is 12.0 N. Therefore, the net force is (75.0 N + 90.0 N) - 12.0 N = 153.0 N.

Next, we can use Newton's second law, F = ma, where F is the net force and m is the mass of the child plus wagon system (23.0 kg). Rearranging the formula to solve for acceleration, a = F/m, we find a = 153.0 N / 23.0 kg = 6.65 m/s².