If a block moving up an inclined.plane at 30° with a velocity of 5m/s , then coefficient of friction will be nearly
Answers
Answered by
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
Answered by
0
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
Similar questions
India Languages,
8 months ago
Computer Science,
8 months ago
Math,
1 year ago
Math,
1 year ago
English,
1 year ago