A person is dragging a packing crate of mass 74.9 kg across a rough horizontal floor where the coefficient of kinetic friction is 0.35. He exerts a force F at and angle 43.0 degrees above the horizontal. What is the Force F such that the crate moves at a constant speed

351.28 N

Explanation:

Let F be the force on the object and f be the frictional force. The component of the force acting in the horizontal direction causing the object to move is FcosФ where Ф is the angle between F and the horizontal = 43.0°. The frictional force on the packing crate f = μN where μ = coefficient of kinetic friction = 0.35 and N = normal force = W = weight of the packing crate = mg where m = mass of crate = 74.9 kg and g = acceleration due to gravity = 9.8 m/s².  So, f = μN = μW = μmg

So, the net force on the packing crate is

FcosФ - f = ma

FcosФ - μmg = ma

Since the crate moves at constant speed, its acceleration a = 0

So, FcosФ - μmg = ma

FcosФ - μmg = m(0)

FcosФ - μmg = 0

FcosФ = μmg

F = μmg/cosФ

Substituting the values of the variables into the equation, we have

F = μmg/cosФ

F = 0.35 × 74.9 kg × 9.8 m/s²/cos43.0°

F = 256.907 kg-m/s²/0.73135

F = 351.28 kg-m/s²

F = 351.28 N