Answer:
a) F = 353.2 N
b) [tex]F_{f} = 147.2 N [/tex]
a = 4.21 m/s²
Explanation:
a) The minimum force necessary to start moving the box s given by:
[tex]\Sigma F = 0[/tex]
[tex] F - \mu_{s}N = 0 [/tex]
[tex] F = \mu_{s}mg [/tex]
Where:
F: is the force applied to move the box
μs: is the static coefficient of friction = 0.6
m: is the mass = 60 kg
g: is the gravity = 9.81 m/s²
[tex] F = 0.6*60 kg*9.81 m/s^{2} = 353.2 N [/tex]
b) The acceleration is:
[tex] F - F_{f} = ma [/tex]
[tex] F - \mu_{k}mg = ma [/tex]
[tex] a = \frac{F - \mu_{k}mg}{m} = \frac{400 N - 0.25*60 kg*9.81 m/s^{2}}{60 kg} = 4.21 m/s^{2} [/tex]
Now, the friction force is:
[tex] F_{f} = \mu_{k}mg = 0.25*60 kg*9.81 m/s^{2} = 147.2 N [/tex]
I hope it helps you!
