Question

A box of mass 60 kg is at rest on a horizontal floor that has a static coefficient of friction of 0.6 and a kinetic coefficient of 0.25
a)The minimum force necessary to start moving the box
b) The friction force and the acceleration of the box if a horizontal force of 400 N is applied

Answer 1

Answer:

a) F = 353.2 N

b) $F_{f} = 147.2 N$

   a = 4.21 m/s²

Explanation:

a) The minimum force necessary to start moving the box s given by:       

$\Sigma F = 0$

$F - \mu_{s}N = 0$

$F = \mu_{s}mg$

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

$F = 0.6*60 kg*9.81 m/s^{2} = 353.2 N$

b) The acceleration is:

$F - F_{f} = ma$

$F - \mu_{k}mg = ma$              

$a = \frac{F - \mu_{k}mg}{m} = \frac{400 N - 0.25*60 kg*9.81 m/s^{2}}{60 kg} = 4.21 m/s^{2}$

Now, the friction force is:

$F_{f} = \mu_{k}mg = 0.25*60 kg*9.81 m/s^{2} = 147.2 N$

I hope it helps you!