A block of 6 kg is pushed across a table by a force P. The block has an acceleration of 4.0m/s^2.A. What is the magnitude and direction of the net force acting upon the block?B. If the magnitude of P is 30n , what is the magnitude and direction of the friction force acting upon the block?

Respuesta :

A.

The net force can be calculated by the formula below:

[tex]F=m\cdot a[/tex]

Where F is the force (in Newtons), m is the mass (in kg) and a is the acceleration (in m/s²).

So we have:

[tex]\begin{gathered} F=6\cdot4 \\ F=24\text{ N} \end{gathered}[/tex]

The force is horizontal (to the right or positive x-axis) and has a magnitude of 24 N.

B.

The friction force has an opposite direction to the applied force, so the net force is given by the difference of forces.

So, calculating the friction force, we have:

[tex]\begin{gathered} F_{\text{net}}=F_P-F_{\text{friction}} \\ 24=30-F_{\text{friction}} \\ F_{\text{friction}}=30-24=6\text{ N} \end{gathered}[/tex]

The friction force is horizontal (to the left or negative x-axis) and has a magnitude of 6 N.