Question

If a block moving up an inclined.plane at 30° with a velocity of 5m/s , then coefficient of friction will be nearly

Answer 1

initial velocity u=5m/s, final velocity v=0

retardation=am/s2,. time=0.5sec

v=u+at

0=5+a×0.5

a=-10m/s2

mg can be resolved as mgcos30° (acting opposite to normal reaction)and mgsin30°.(acting downwards)

(R ) normal reaction=mgcos30°

friction f=uR, where u is cofficient of friction

net force= -(f+ mgsin30°)

ma=-(uR+mgsin30)

m(-10)=-(u.mgcos30+mgsin30)

10m=m(u×√3/2+1/2).g

10/g=√3u/2+1/2

take g=10, and solving the equation

u=1/√3

Answer 2

Initial Velocity u = 5 m/s

Time Taken t = 0.5 s

Final Velocity = 0

v = u + at

0 = 5 + 0.5a

a = - 10m/s^2

This stopping force is due to the frictional force.

fr = umg cos θ

ma = umg cos θ

u = a/g cos

u = 10/10 cos 30

u = 2/√3

u = 1.15