The conveyer
belt shown in the figure is used for material handling,where the weight
of each package is 15 kg.The coefficient of friction is mk
=0.15.Calculate the distance the package will cover when it is brought to
a sudden halt from a speed of 5 m/s. a)8.4947 m b)9.487 m c)10.47 m d)6.487 m. 

Free body
diagram The frictional force acting on the pack is directed opposite to its direction of motion and is given by F= 0.15 Nc where Nc is the normal reaction on the pack given by the conveyer belt. Equations of motion The acceleration a is given by a = 1.4715 m/s2. Kinematics v2 = u2 + 2 as, 0 = 25 + 2*(1.4715)s, s = 8.4947 m. 