Two blocks rest on a horizontal frictionless surface. (top block weighs xkg, the

Question

Two blocks rest on a horizontal frictionless surface. (top block weighs xkg, the bottom box weighs 10kg). The surface between the top and bottom blocks is roughened so that there is no slipping between the two blocks. A 40N force is applied to the bottom block and the two blocks are found to be moving together with an acceleration a=2m/s^s

(a) what is the mass of the top box?

     a.1kg   b.2kg    c.5kg    d.10kg    e.15kg

(b) what is the force of static friction between top and bottom blocks

      a.0N  b.10N  c.20N  d.25N  e.30N

(c) what is the minimum coefficient of static friction necessary to keep the tip block from slipping on the bottom block.

     a. 0.05  b. 0.10  c. 0.20  d.0.30  e.0.40

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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