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Block B has mass 6.00 kg and sits at rest on a horizontal, frictionless surface. Block A has mass 2.50 kg and sits at rest on top of block B. The coefficient of static friction between the two blocks is 0.400. A horizontal force P is then applied to block A. What is the largest value P can have and the blocks move together with equal accelerations?

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Answer:

Explanation:

Block A sits on block B and force is applied on block A . Block A will experience two forces 1) force P and 2 )  friction force in opposite direction of motion . Block B will experience one force that is force of friction in the direction of motion .

Let force on block A be P . friction force on it will be equal to kinetic friction, that is μ mg , where μ is coefficient of friction and m is mass of block A

friction force = .4 x 2.5 x 9.8

= 9.8 N

net force on block A = P - 9.8

acceleration = ( P - 9.8 ) / 2.5

force on block B = 9.8

acceleration = force / mass

= 9.8 / 6

for common acceleration

( P - 9.8 ) / 2.5  = 9.8 / 6

( P - 9.8 ) / 2.5 = 1.63333

P = 13.88 N .

pstnonsonjoku

The acceleration is 3.722 ms-2.

For the block B;

0.400 = F/6.00 kg × 9.8 ms-2

F = 0.400 × 6.00 kg × 9.8 ms-2

F = 23.52 N

For block A;

0.400 = F/2.50 kg × 9.8 ms-2

F = 0.400 × 2.50 kg × 9.8 ms-2

F = 9.8 N

For common acceleration; 23.52 N - 9.8 N/(6 Kg - 2.5 Kg) =

a = 3.722 ms-2

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