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

explanation:

considering the bohr theory, the type of connection that requires the transfer the answer conuldn't be right

when two balls of the different masses are dropped from equal height, they reach the ground at the same time .this is because they experience the same acceleration.

the acceleration is independent of mass.

thus when they are dropped they will always maintain the same velocity and travel the same distance (neglect air resistance of course) at the same time.