150-N box is being pulled horizontally in a wagon accelerating uniformly at 3.00 m/s2. The box does not move relative to the wagon, the coefficient of static friction between the box and the wagon's surface is 0.600, and the coefficient of kinetic friction is 0.400. The friction force on this box is closest to

the friction force on this box is closest to 45.9 N.

Explanation:

given data

Weight of the box W = 150 N

accelerating uniformly = 3.00 m/s²

coefficient of kinetic friction = 0.400

coefficient of static friction = 0.600

solution

we know box does not move relative to the wagon

we get here friction force  that is express as

friction force = mass × acceleration ..............1

here mass = weight ÷ g

mass = = 15.3 kg

put value in equation 1 we get

friction force = 15.3 kg × 3

friction force =  45.9 N

So, the friction force on this box is closest to 45.9 N.

u=13

v=30

s/t=u+1\2at

=13+431/2

=13+265.5

=278.5

positive reinforcement? ?