Two blocks are placed on a frictionless inclined plane as shown in the figure. T

Question

Two blocks are placed on a frictionless inclined plane as shown in the figure. There is friction between the blocks. The coefficient of kinetic friction is Uk= 0.14 and that of static friction is Us=0.55. A string is tied to the lower block, and the blocks are pulled up the plane with a force P= 31.5 N due to which the blocks accelerate with an acceleration of a= 0.667 m/s^2. If the block m1=2kg does not slip on the block m2=5 kg while the blocks are accelerating together, what will be the magnitude of static friction force between the blocks m1 and m2? Theta is given as 23 degrees

I got 9.9N but they are saying answer is 9

Explanation / Answer

accleration of 2kg block is 0.667 m/s^2

So applying newton law

Friction force - component of gravitation force = mass*accleration

=> cofficient_fri*m*g*cos - m*g*sin = m*a

=> magnitude of static friction force = m*a + m*g*sin = 2*0.667 + 2*9.8*0.39 = 8.978N

= 9 N Answer

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