A body of mass 1kg lies on an inclined plane of angle 60 to the horizantal. If the coefficient of friction is 0.4 the frictional force along the inclined plane is
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2
Hey Mate
Limiting friction, f = μmg = 0.6 × 1 × 9.8 = 5.88 N
Applied force = F = ma = 1 × 5 = 5 N
As F < f, so force of friction = 5 N
hope this answer helps you
please mark it as brainliest
Limiting friction, f = μmg = 0.6 × 1 × 9.8 = 5.88 N
Applied force = F = ma = 1 × 5 = 5 N
As F < f, so force of friction = 5 N
hope this answer helps you
please mark it as brainliest
Answered by
5
HIII ......
GOOD MORNING;-)
HERE IS YOUR ANSWER:--------
f= 0.6×1×9.8= 5.88N.
APLLIED FORCE= 1×5=5N
THEREFORE ,5N IS THE ANSWER.
HOPE THIS HELPS YOU
BE ALWAYS HAPPY AND SMILING.......
GOOD MORNING;-)
HERE IS YOUR ANSWER:--------
f= 0.6×1×9.8= 5.88N.
APLLIED FORCE= 1×5=5N
THEREFORE ,5N IS THE ANSWER.
HOPE THIS HELPS YOU
BE ALWAYS HAPPY AND SMILING.......
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