A block of mass 4.0 kg is put on top of a block of mass M = 6.0 kg. To cause the

Question

A block of mass 4.0 kg is put on top of a block of mass M = 6.0 kg. To cause the top block to slip on the bottom one, while the bottom one is held fixed, a horizontal force of at least 11 N must be applied to the top block. The assembly of blocks is now placed on a horizontal, frictionless table.
(a) Find the magnitude of the maximum horizontal force F that can be applied to the lower block so that the blocks will move together.
N

(b) Find the magnitude of the resulting acceleration of the blocks.
m/s2

Explanation / Answer

When the bottom block is fixed : force(frictional) to move top block, F = 19N therefore, uN = 19 N- normal reaction between blocks = mg = 4*9.81= 39.24 When both blocks are moved applying force on bottom block: forces on bottom block : F from top block in reverse direction , since ground is frictionless no force from ground. therefore, force required = 19 N with acceleration = 19/(4 + 5.5) = 2 m/s2

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