Question

2 kg mass starts from rest on an inclined smooth surface with inclination 30o and length 2 m. how much will it travel before coming to rest on a frictional surface with frictional coefficient of 0.25

Answer 1

After travelling a distance of 2 meters

The final Velocity becomes,

v^2 = 0^2 + 2gsin30*2

v = root(20) = m/s.

 

From friction surface starts, we would apply WET,

Hence,

-(0.25*2*g*cos30)*x  = 0 –  0.5*2*v^2  { friction force is negative}

put v value and solve we get,

x = 2.312 meter.