Question

Suppose two children push horizontally, but in exactly opposite directions, on a wagon containing a third child. The first child exerts a force of F1 = 75.0 N, the second a force of F2 = 90.0 N, friction is 12.0 N, and the mass of the third child plus wagon is 30.0 kg. Calculate the acceleration (in m/s2)

Answer 1

Answer:

$0.1 m/s^2$, in the direction of F2

Explanation:

The magnitude of F2 is larger than that of F1: this means that the wagon will tend to move in the direction of F2. Therefore, the direction of friction will be opposite to the direction of motion, therefore in the direction of F1.

Let's now take the direction of the force exerted by the second child, F2, as positive direction. Then the direction of F1 and of the force of friction ($F_f$) will be negative: so we have

$F_2 = +90.0 N\\F_1 = -75.0 N\\F_f = -12.0 N$

We can write the equation of the forces for the third child+wagon:

$F_1+F_2+F_f = ma$

where

m = 30.0 kg is the combined mass of the child + the wagon

Now we can solve the equation for a, to find the acceleration:

$a=\frac{F_2+F_1+F_f}{m}=\frac{90-75-12}{30}=0.1 m/s^2$