Question

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

Answer 1

5.4 N

Step-by-step 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

Answer 2

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 =
`1.5m/s^(2)`

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:


`Tcos(45)-f_(k)=ma_(x)`

Isolating
`f_(k)`:


`f_(k)=Tcos(45)-ma_(x) = 50cos(45)-20(1.5)=5.355N`

That is the friction force on the box.