Astudent pulls on a 20 kg box with a force of 50 n at an angle 45 degrees relative to the horizontal. the box increases speed at a rate of 1.5 m/s2. what is the approximate value of the friction force on the box

For this problem the figure below shows the representation of a student who pulls on a 20kg box. We know this variables:

Weight of the box = 20kg
Force used by the student to pull on the box = 50N (This is the tension T)
Angle relative to the horizontal = 45 degrees
Aceleration of the box = 

The figure also shows the Free-Body diagram, Applying Newton's Second Law we can find the equation for this diagram, related to the x-axis as:



Isolating :



That is the friction force on the box.

We don't know . . .

-- the speed of the airplane at the moment he jumped

-- the altitude of the airplane at the moment he jumped

-- the speed or direction of the winds at every altitude all the way down

-- the position he was in during the 4 seconds of "free"-fall

-- the skydiver's weight

-- the drag of his parashoot

Otherwise, this sounds like a fun problem to work on.

I can tell you this:  

The vertical component of the skydiver's total fall is equal to

(the airplane's elevation above sea level at the moment he jumps)

minus

(the elevation above sea level of the ground upon which he lands).