A horizontal force of 3.2 N is applied to a 1.2- kg radio,initially at rest on a
ID: 1681395 • Letter: A
Question
A horizontal force of 3.2 N is applied to a 1.2- kg radio,initially at rest on a table with a level surface.Will the radio move if the coefficient of friction is 0.4 (One tryonly !!!)?
What is the coefficient of friction if itjust begins to move?
What is its acceleration if the coefficient offriction is only 0.13? A horizontal force of 3.2 N is applied to a 1.2- kg radio,initially at rest on a table with a level surface.
Will the radio move if the coefficient of friction is 0.4 (One tryonly !!!)?
What is the coefficient of friction if itjust begins to move?
What is its acceleration if the coefficient offriction is only 0.13?
Explanation / Answer
now here normal reaction acting = 1.2 x 9.81 = 11.772 N horizontal force = 3.2 N a) now if the coefficent of friction is 0.4 we have frictionalforce = 0.4 x 11.772 = 4.7 N now as the frictional force is more tahn the horizontal force soteh block will not move in case a) b) to make it just move we have coefficent of friction = 3.2/11.772 = 0.271 or less than that. c)now here we have acceleration a = (F - frictional force )/m =( 3.2 - 11.772 x 0.13)/ 1.2 = 1.39 m/s2
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