Figure shown two blocks in contact sliding down an inclined surface of inclination `30^(@)`. The block of mass `2 kg `and the incline is `mu_(1) = 0.20` and the incline is `mu_(2) = 0.30` .Find the acceleration of `2.0 kg` block (in `ms^(-2)`)`g = 10 ms^(-2)`
Text Solution
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Since ,`mu_(1) lt mu_(2)` acceleration of `2 kg` block down drow the plane will be more than the acceleration of `4 -kg` block ,if allowed to move sepseparatcly . In this , both blocks are treated as a system of mass `(4 + 2) = 6` kg and will move down with the same acceleration . Net force down the plane is `F = (m_(1) + m_(2)) g sin theta - mu_(1) m_(1) g cos theta - mu_(2) m_(2) g cos theta` `=(4 + 2)g sin 30^(@) - (0.2)(2) g cos 30^(@)- (0.3) (4) g cos 30^(@)` ` = (6)(10) ((1)/(2)) - (0.4)(10) (sqrt(3)/(2)) - (1.2)(10) (sqrt(3)/(2))` `= 30 - 13.76 = 16.24 N` Therefore , acceleration of both the blocks down the plane will be `a = (F)/(m_(1) + m_(2)) = (16.24)/(4 +2) = 2.7 ms^(-2)`
Figure shown two blocks in contact sliding down an inclined surfase of inclination 30^(@) . The friction coefficient between the block of mass 4.0kg and the incline is mu _(2) =0.30 . Find the acceleration of 2.0kg block. (g = 10m//s^(2))
Figure shows two blocks in contact placed on an incline of angle theta = 30^(@) . The coefficient of friction between the block of mass 4 kg and the incline is m1, and that between 2 kg block and incline is mu_(2) . Find the acceleration of the blocks and the contact force between them if – (a) mu_(1) = 0.5 , mu_(2) = 0.8 (b) mu_(1) = 0.8 , mu_(2) = 0.5 (c) mu_(1) = 0.6 , mu_(2) = 0.1 " " [ Take g = 10m//s^(2) ]
A block slides down on inclined plane (angle of inclination 60^(@) ) with an accelration g//2 Find the coefficient friction
Two blocks 4 kg and 2 kg are sliding down an incline plane as shown in figure. The acceleration of 2 kg block is.
Two blocks 4 kg and 2 kg are sliding down an incline plane as shown in fig The acceleration of 2 kg block is:
A block of mass 5 kg slides down a rough inclined surface. The angle of inclination is 45^(@) . The coefficient of sliding friction is 0.20. When the block slides 10 cm, the work done on the block by force of friction is
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Two blocks m_(1) = 4 kg and m_(2) = 2 kg connected by a weightless rod slide down a plane having an inclination of 37^(@) . The coefficient of dynamic friction of m_(1) and m_(2) with the inclined plane are mu_(1) = 0 .75 and mu_(2) = 0 . 25 respectively Find the common acceleration of the two blocks and tension in the rod Take is 37^(@) = 0.6 and cos 37^(@) = 0 .8 .
CENGAGE PHYSICS-NEWTON'S LAWS OF MOTION 2-Integer type
