Answer to Question #239482 in Physics for rshiii

Question

6. A stockroom worker pushes a box with a mass of 11.2 kg on a horizontal surface with a constant speed of 3.5 m/s. The coefficient of kinetic friction between the box and the surface is 0.20. a) What horizontal force must be applied by the worker to maintain the motion? b) If the force calculated in part (a) is removed, how far does the box slide before coming to rest? 

Answer 1

Given:

$m=11.2\:\rm kg$

$v_0=3.5\:\rm m/s=const$

$\mu=0.20$

The Newton's second law says

$$ma=F-F_f$$

(a) $a=0$

$$F=F_f=\mu N=\mu mg=0.2*11.2*9.8=22\:\rm N$$

(b) $F=0$

$$ma=-F_f=-\mu mg$$

$$a=-\mu g=-0.2*9.8=-1.96\:\rm m/s^2$$

Distance

$$d=\frac{v^2-v_0^2}{2a}=\frac{0^2-3.5^2}{2*(-1.96)}=3.1\:\rm m$$