Two blocks rest on a horzontal frictionless surface. The surface between the top

Question

Two blocks rest on a horzontal frictionless surface. The surface between the top and botom blocks is roughened so that is no slipping between the two blocks. A 40 N force is applied to the botom block and the two blocks are found to be moving togther with an acceleration a=2m/s^2.

1) What is the value of mass of the upper block?

2) What is the force of stati friction between the top and bottom blocks?

3) What is the maximum coefficient of static friction necessary to keep the top block from slipping on the bottom block?

Please show full explanation and steps for answers.

Explanation / Answer

let the force of friction between the two blocks be f N

acceleration a = 2 m/s^2

a = 40 / ( x+10 )

x = 10 kg

(d) Mass of top box is 10kg

acceleration of top block is 2m/s^2 [ as both blocks are moving together both have the same acceleration ]

and net force actiong on top block is only the force of static friction f

So, f/10 = 2 ==> f = 20 N

(c) Force of static friction betwen two blocks is 20N

normal force acting on the top block = mg

friction force = f = umg where u is the co-efficient of static friction

for the blocks to not slip we must have cceleration of first block and second block to be equal

(F - f) / M = f /m

F is external force

f is frictional force

M is mass of bottom block and m is mass of top block

Solving the equation we have u = 0.20

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