A block of weight `100N` lying on a horizontal surface just beging to when a horizontal to move when a horinzontal force of `25 N` acts on it Determine the coefficent of static friction
Text Solution
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As the `25- N` force brings the block to the point of sliding , the frictional force `= mu _(s) N` from force diagram `n = 100 n` `mu _(s) N` `rArr mu _(s)= 0.25`
A block of weight 200N is pulled along a rough horizontal surface at contant speed by a force of 100N acting at an angle 30^(@) above the horizontal. The coefficient of kinetic friction between the block and the surface is .
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A block of mass 10kg is placed on a rough horizontal surface having coefficient of friction mu=0.5 . If a horizontal force of 100N is acting on it, then acceleration of the will be.
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A 10-kg block is placed on a horizontal surface. The coefficient of friction between them is 0.2 . A horizontal force P = 15 N first acts on it in the eastward direction. Later, in addition to P a second horizontal force Q = 20 N acts on it in the northward direction.
A horizontal force of 20 N is applied to a block of 10 kg resting on a rough horizontal surface. How much additional force is required to just move the block, if the coefficient of static friction between the block and the surface is 0.4 ?
A block of mass 20 kg is placed on a rough horizontal surface. When a force of 80 N is applied at an angle of 30°with the horizontal, the block just begins to slide. What is the coefficient of static friction ? (g= 10m//s^2 )
A 10 kg block is placed on a horizontal surface whose coefficient of friction is 0.2.A horizontal force P = 15 N first acts on it in the eastward direction. Later, in addition to P a second horizontal force Q = 20 N acts on it in the northward direction: (Take g= 10m/s^(2) )
CENGAGE PHYSICS-NEWTON'S LAWS OF MOTION 2-Integer type
