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

Answer 1

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, $F cos \theta$, forward
• The friction force, $F_f$, backward

The resultant force must be equal to the product between mass and acceleration (Newton's second law), so we can write:

$F cos \theta - F_f = ma$

where:

F = 50 N is the force of pull of the student

$\theta=45^{\circ}$ is the angle relative to the horizontal

m = 20 kg is the mass of the box

$a=1.5 m/s^2$ is the acceleration

Solving the equation for $F_f$, we find the value of the friction force:

$F_f = Fcos \theta -ma=(50)(cos 45^{\circ})-(20)(1.5)=5.4 N$

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