The man pushes with a force of 200 N. The child and sled combo has a mass of 30 kg and the coefficient of static friction is 0.15. Determine the normal force between sled and the plane.
The first step is to determine the weight of the child and sled combo, which is given by:
w = mg = 30 kg x 9.81 m/s^2 = 294.3 N
The normal force (N) between the sled and the plane is equal and opposite to the weight, since the sled is not accelerating in the vertical direction:
N = w = 294.3 N
The maximum force of static friction (fs) that can be applied between the sled and the plane is given by:
fs = μsN
where μs is the coefficient of static friction. Therefore:
fs = 0.15 x 294.3 N = 44.145 N
Since the man is pushing with a force of 200 N, which is greater than the maximum force of static friction, the sled will move with an acceleration:
a = F/m = (200 N - fs)/30 kg = (200 N - 44.145 N)/30 kg = 5.185 m/s^2
Note that the force of kinetic friction (fk) will be smaller than the force of static friction. Once the sled starts moving, the coefficient of kinetic friction (μk) will apply, which is generally smaller than the coefficient of static friction. If we assume that the coefficient of kinetic friction is 0.10, the force of kinetic friction will be:
fk = μkN = 0.10 x 294.3 N = 29.43 N
Therefore, the man will have to continue applying a force of at least 29.43 N to keep the sled moving at a constant speed after it starts moving.