A child pulls a 13.4 kg wagon across a level floor, causing it to accelerate at 6.2 m/s^2. If the force of friction between the wheels and the wagon is 5.7 N, what is the force applied by the child?

Answer 1

Equate 13.4 kilograms times 6.2 meters per second-squared to 83.08 kilogram • meters per second-squared (or Newtons).

The Force applied by the child is then found by (83.08 + 5.7) Newtons or 88.78 Newtons.

The child must supply the Force to reach 83.08 Newtons as well as an additional 5.7 Newtons to "zero" or cancel the retarding friction.

Answer 2

Considering ∑F = ma, we first add the opposing forces together:

F_A - F_f = ma,

where F_A is the force of the child pulling the wagon, F_f is the frictional force.

Then we simply solve for F_A:

F_A = ma + F_f,

where all variables on the right-hand side are known.