ch06-p034

ch06-p034 - 34. The free-body diagrams for the slab and...

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which is greater than f s ,max so that we conclude the block is sliding across the slab (their accelerations are different). (a) Using f = μ k Nb F the above equations yield 2 2 (0.40)(10 kg)(9.8 m/s ) 100 N 6.1 m/s . 10 kg kb b b mg F a m == = The negative sign means that the acceleration is leftward. That is, 2 ˆ ( 6.1 m/s )i b a =− G 34. The free-body diagrams for the slab and block are shown below. G F is the 100 N force applied to the block, Ns F G is the normal force of the floor on the slab, F is the magnitude of the normal force between the slab and the block, G f is the force of friction between the slab and the block, m s is the mass of the slab, and m b is the mass of the block. For both objects, we take the + x direction to be to the right and the + y direction to be up. Applying Newton’s second law for the

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This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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ch06-p034 - 34. The free-body diagrams for the slab and...

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