A man of mass 70 kg is standing stationary with respect to a horizontal conveyor belt that is accelerating at `1.5ms^( - 2)`. If the coefficient of friction between man's shoes and belt is 0.2, the frictional force on the man is (`g = 10ms^( - 2)`) (A) 245 N (B) 700 N (C) 140 N (D) 105 N

Figure shows a man of mass 55 kg standing stationary with respect to a horizontal conveyor belt that is accelerating with 1m s^(-2) . The net force acting on the man is

Figure shows a man standing stationary with respect to a horizontal conveyer belt that is accelerating with 1 ms^(-2) What is the net force on the man? (Mass of the man = 65 kg)

Figure shown a man standing stationary with respect to a horizontal converyor belt that is accelerationg with 1m//s^(-2) . What is the net force on the man?If the coefficient of ststic friction between the man's shoes and the belt is 0.2 upto what maximum acceleration of the belt can the man continue to be stationary relative to the belt? Mass of the man = 65kg (g =9.8m//s^(2))

A man of mass 65kg is standing stationary with respect to a conveyor belt which is accelerating with 1m//s^(2) . if mu_(s) is 0.2 the net force on the man and the maximum acceleration of the belt so that the man is stationary relative to the belt are (g =10 m//s^(2)) .

A block of mass 1 kg is placed on a truck which accelerates with acceleration 5ms^(-2) . The coefficient of static friction between the block and truck is 0.6. The frictional force acting on the block is

A block of mass 1kg is pressed against a wall by applying a horizontal force of 10N on the block. If the coefficient of friction between the block and the wall is 0.5 ,magnitude of the frictional force acting on the block is

A block of mass 0.1 kg is held against a wall by applying a horizontal force of 5 N on the block. If the coefficient of friction between the block and the wall is 0.5, then the magnitude of the frictional force acting on the block is