Given,
The weight of the box, W=300 N
The coefficient of static friction, μ=0.30
(a) In order to move the box, the force applied must be greater than the static friction that exists between the box and the floor. The static friction is given by,
[tex]\begin{gathered} f=N\mu \\ =W\mu \end{gathered}[/tex]
Where N is the normal force which is the same as the weight of the box.
On substituting the known values,
[tex]\begin{gathered} f=300\times0.3 \\ =90\text{ N} \end{gathered}[/tex]
Thus the horizontal force required to start the box into motion is 90 N.
(b) The applied force is F= 50.0 N
The coefficient of kinetic friction μ=0.3
Thus, the kinetic friction offered by the floor is,
[tex]f_k=W\mu[/tex]
On substituting the known values,
[tex]f_k=300\times0.3=90\text{ N}[/tex]
Thus the net force acting on the body will be,
[tex]\begin{gathered} F_n=ma \\ =F-f_k\text{ }\rightarrow(i) \end{gathered}[/tex]
Where m is the mass of the box and a is the acceleration of the box
The mass is given by
[tex]m=\frac{W}{g}[/tex]
Where g is the acceleration due to gravity.
On substituting the known values,
[tex]\begin{gathered} m=\frac{300}{9.8} \\ =30.6\text{ kg} \end{gathered}[/tex]
Thus, from equation (i) the acceleration is calculated as
[tex]\begin{gathered} a=\frac{F-f_k}{m} \\ =\frac{50-90}{30.6} \\ =-1.31m/s^2 \end{gathered}[/tex]
Thus the acceleration of the box will be -1.31 m/s². The negative sign indicates that it is a deceleration. Thus, the box will eventually slow down and will come to rest.
