The acceleration of the box that has a mass of 4 kg and a coefficient of kinetic friction of 0.2, when moved by an exerted force of 58 N, is 10.06 m/s².
The acceleration of the box can be calculated with the sum of the forces acting on the x-direction:
[tex] \Sigma F_{x} = ma [/tex]
We will not take into account the forces acting on the y-direction since we need to find the acceleration of the box when it is moving (x-direction).
[tex] T_{x} - F_{\mu} = ma [/tex] (1)
Where:
From equation (1) we have:
[tex] Tcos(34) - \mu N = ma [/tex]
[tex] Tcos(34) - \mu mg = ma [/tex]
Hence, the acceleration is:
[tex] 58N*cos(34) - 0.2*4 kg*9.81 m/s^{2} = 4 kg*a [/tex]
[tex] a = \frac{58N*cos(34) - 0.2*4 kg*9.81 m/s^{2}}{4 kg} = 10.06 m/s^{2} [/tex]
Therefore, the acceleration of the box when the box is moving is 10.06 m/s².
You can learn more about forces here:
I hope it helps you!