How much work did the movers do (horizontally) pushing a 43.0-kg crate 10.4 m across a rough floor without acceleration, if the effective coefficient of friction was 0.60

The crate is in equilibrium. Newton's second law gives

∑ F (vertical) = n - mg = 0

∑ F (horizontal) = p - f = 0

where

• n = magnitude of the normal force

• mg = weight of the crate

• p = mag. of push exerted by movers

• f = mag. of kinetic friciton, with f = 0.60n

It follows that

p = f = 0.60mg = 0.60 (43.0 kg) g = 252.84 N

so that the movers perform

W = p (10.4 m) ≈ 2600 J

of work on the crate. (The total work done on the crate, on the other hand, is zero because the net force on the crate is zero.)

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air exerts forces on falling objects near earth’s surface.

explanation:

the term 'free falls' refer to object that are falling towards the earth, in a situation in which gravity is the only force acting on the object.

for an object in free fall, the acceleration is constant (g, the gravitational acceleration), and so the velocity of the object constantly increases.

for objects near earth's surface, there is always another force acting on the object: the resistance due to the air. this force is opposed to the direction of fall of the object, and it is proportional to the velocity of the object, so as the object falls down, this force increases up to a point when it balances the force of gravity, and the object continues its motion at constant velocity (called terminal velocity). since in this case gravity is not the only force acting on the object, we are not in a situation of free fall.