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.
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.
v2 = u2 + 2 as,
0 = 25 + 2*(-1.4715)s,
s = 8.4947 m.