if a block moving up an inclined plane at 30 with a velocity of 5 m/s , stops after 0.5 seconds then coeficient of friction will be nearly
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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
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
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