A student 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 = [tex]1.5m/s^{2} [/tex]

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:

[tex]Tcos(45)-f_{k}=ma_{x}[/tex]

Isolating [tex]f_{k}[/tex]:

[tex]f_{k}=Tcos(45)-ma_{x} = 50cos(45)-20(1.5)=5.355N[/tex]

That is the friction force on the box.

yakndaraudoimeh yakndaraudoimeh · Answer 2 · 2021-02-19T07:05:20+01:00

yakndaraudoimeh yakndaraudoimeh

19-02-2021

Answer:

5.4 N

Explanation

sin 45 = x/50

sin 45 * 50 = 35.4N (upward)

50^2= 35.4^2 + c^2

c^2 = 1246.84

c = 35.3N (X component)

1.5m/s^2 * 20kg = 30

35.3 - 30 = 5.4

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