Question

A block of mass 1 kg is at rest on a horizontal table the coefficient of friction between block and the table is 0.50 is equal to 10 metre per second and the magnitude of a force acting upwards at an angle of 60° from the horizontal that will start block

Answer 1

The magnitude of  force acting upward on the block is F = 20 / 2+ √ 3

Explanation:

R+ Fsin(60°) = mg

R=mg − √3F / 2 ----(2)

If block just starts moving

F cos(60°) = f = μR

F+ √3F / 2 = 10

F = 20 / 2+ √ 3

Thus the magnitude of  force acting upward on the block is F = 20 / 2+ √ 3

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