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

35 N

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45 N

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55 N

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65 N

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Solution

The correct option is C
55 N

As the man is standing stationary w.r.t. the belt, acceleration of man = acceleration of belt

$∴$ Net force on the man

$F = M a = (55 k g) (1 m s^{- 2}) = 55 N$

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Figure 5.18 shows a man standing stationary with respect to a horizontal conveyor belt that is accelerating with $1 m s^{- 2}$ . What is the net force on the man? If the coefficient of static friction between the man's shoes and the belt is 0.2, up to what acceleration of the belt can the man continue to be stationary relative to the belt? (Mass of the man = 65 kg.)