The value of the friction force is 5.4 N
Explanation:
In order to solve the problem, we have to write down the equation of the forces along the horizontal direction.
We have only two forces acting along the horizontal direction:
- The horizontal component of the pull of the student, [tex]F cos \theta[/tex], forward
- The friction force, [tex]F_f[/tex], backward
The resultant force must be equal to the product between mass and acceleration (Newton's second law), so we can write:
[tex]F cos \theta - F_f = ma[/tex]
where:
F = 50 N is the force of pull of the student
[tex]\theta=45^{\circ}[/tex] is the angle relative to the horizontal
m = 20 kg is the mass of the box
[tex]a=1.5 m/s^2[/tex] is the acceleration
Solving the equation for [tex]F_f[/tex], we find the value of the friction force:
[tex]F_f = Fcos \theta -ma=(50)(cos 45^{\circ})-(20)(1.5)=5.4 N[/tex]
Learn more about friction:
brainly.com/question/6217246
brainly.com/question/5884009
brainly.com/question/3017271
brainly.com/question/2235246
#LearnwithBrainly
