Aman exerts a constant force to pull a 49-kg box across a floor at constant speed. he exerts this force by attaching a rope to the box and pulling so that the rope makes a constant angle of 36.9∘ above the horizontal. the coefficient of kinetic friction for the box-floor interface is μk = 0.20.

what is the work done by the man as he moves the box 10 m?

the velocity is constant so fnet=0 and
f horizontal=f friction
f horizontal=f man × cos36.9
f friction=0.2(9.8×49 - f up)
f up = f man × sin36.9

so
fm(cos36.9)=0.2(9.8×49 - fm(sin36.9))
solve for fm: 104.4 n

multiply by 10 to get 1044 that's wrong.

840 J

Explanation:

In the y direction:

∑F = ma

Fn + F sin θ − mg = 0

Fn = mg − F sin θ

In the x direction:

∑F = ma

F cos θ − Fn μ = 0

F cos θ = Fn μ

F cos θ = (mg − F sin θ) μ

F cos θ = mg μ − F μ sin θ

F (cos θ + μ sin θ) = mg μ

F = mg μ / (cos θ + μ sin θ)

F = (49) (9.8) (0.2) / (cos 36.9° + 0.2 sin 36.9°)

F = 104.4 N

You correctly found the force applied by the man, but remember that only the horizontal component of that force is used to do work.

W = F cos θ d

W = (104.4) (cos 36.9°) (10)

W = 835

Rounding to two significant figures, the man does 840 J of work.

d. air pressure decreases as altitude increases.

the total distance that martina travels that day is

(distance from home to mom's office) plus (distance from mom's office to school)

plus (distance from school to library) plus (distance from library to friend's house)

plus (distance from friend's house to martina's house).

(we don't know what any of these distance actually is.)