Answer:
[tex]1.176 \frac{m}{s^{2}}[/tex]
Explanation:
[tex]m[/tex] = Total mass of carton of books = 120 kg
[tex]\mu _{s}[/tex] = Coefficient of static friction = 0.42
[tex]\mu _{k}[/tex] = Coefficient of kinetic friction = 0.30
[tex]F[/tex] = force applied to get the box moving
[tex]f[/tex] = kinetic frictional force acting on the carton of books
[tex]a[/tex] = acceleration of the box
Force applied to get the box moving is given as
[tex]F = \mu _{s}mg[/tex]
kinetic frictional force is given as
[tex]f = \mu _{k}mg[/tex]
Force equation for the motion of the box of books is given as
[tex]F - f = ma[/tex]
[tex]\mu _{s}mg - \mu _{k}mg = ma[/tex]
[tex]\mu _{s}g - \mu _{k}g = a[/tex]
[tex](0.42)(9.8) - (0.30)(9.8) = a[/tex]
[tex]a = (0.42)(9.8) - (0.30)(9.8)[/tex]
[tex]a = 1.176 \frac{m}{s^{2}}[/tex]
