A BLOCK WEIGHING 10KG IS AT REST ON HORIZONTAL TABLE THE COEFFICIENT FRICTION BETWEEN BLOCK AND TABLE IS 0.5 IF A FORCE ACTS DOWNWARDS AT 60 DEGREE WITH THE HORIZONTAL , HOW LARGE CAN IT TO BE WITHOUT CAUSING THE BLOCK TO MOVE

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Arun · Answer

Dear student a = 0.5 * 10 = 5 m/sec^2 Now F = ma = 10 * 5 = 50N Hence it can be any force till you apply 50N without causing the block to move

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A BLOCK WEIGHING 10KG IS AT REST ON HORIZONTAL TABLE THE COEFFICIENT FRICTION BETWEEN BLOCK AND TABLE IS 0.5 IF A FORCE ACTS DOWNWARDS AT 60 DEGREE WITH THE HORIZONTAL , HOW LARGE CAN IT TO BE WITHOUT CAUSING THE BLOCK TO MOVE

A BLOCK WEIGHING 10KG IS AT REST ON HORIZONTAL TABLE THE COEFFICIENT FRICTION BETWEEN BLOCK AND TABLE IS 0.5 IF A FORCE ACTS DOWNWARDS AT 60 DEGREE WITH THE HORIZONTAL , HOW LARGE CAN IT TO BE WITHOUT CAUSING THE BLOCK TO MOVE

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A BLOCK WEIGHING 10KG IS AT REST ON HORIZONTAL TABLE THE COEFFICIENT FRICTION BETWEEN BLOCK AND TABLE IS 0.5 IF A FORCE ACTS DOWNWARDS AT 60 DEGREE WITH THE HORIZONTAL , HOW LARGE CAN IT TO BE WITHOUT CAUSING THE BLOCK TO MOVE

A BLOCK WEIGHING 10KG IS AT REST ON HORIZONTAL TABLE THE COEFFICIENT FRICTION BETWEEN BLOCK AND TABLE IS 0.5 IF A FORCE ACTS DOWNWARDS AT 60 DEGREE WITH THE HORIZONTAL , HOW LARGE CAN IT TO BE WITHOUT CAUSING THE BLOCK TO MOVE

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