A marble block of mass2kg lying on ice.When given a velocity of 6ms it stopped by friction in 10s.Then the coefficient of friction is
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Mass of block, M = 2 kg
Initial velocity, u = 6m/s
Final velocity , v =0
time ,t = 10 s
a=v−ut=0−610=−0.6 m/s2
Coefficient of friction = a/g
μ=−0.69.8=−0.061
-ve means direction of frictional force is opposite to motion of marble.
Initial velocity, u = 6m/s
Final velocity , v =0
time ,t = 10 s
a=v−ut=0−610=−0.6 m/s2
Coefficient of friction = a/g
μ=−0.69.8=−0.061
-ve means direction of frictional force is opposite to motion of marble.
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Answer:
Explanation:
Mass of block, M = 2 kg
Initial velocity, u = 6m/s
Final velocity , v =0
time ,t = 10 s
a=v−ut=0−610=−0.6 m/s2
Coefficient of friction = a/g
μ=−0.69.8=−0.061
-ve means direction of frictional force is opposite to motion of marble.
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