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Laws of Motion — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsLaws of Motion5 Marks
Question
Figure. shows a man standing stationary with respect to a horizontal conveyor belt that is accelerating with $1ms^{-2}$. 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 = $65kg$.)
✓

Answer

Mass of the man, m = 65kg Acceleration of the belt, $a = 1m/s^2$ Coefficient of static friction, μ = 0.2 The net force F, acting on the man is given by Newton’s second law of motion as: $F_{net} = ma = 65 × 1 = 65N$ The man will continue to be stationary with respect to the conveyor belt until the net force on the man is less than or equal to the frictional force fs, exerted by the belt, i.e., $F’_{net} = f_s ma’ = \mu mg $
​​​​​​​$\therefore a‘ = 0.2 \times 10 = 2m/s^2$ Therefore, the maximum acceleration of the belt up to which the man can stand stationary is $2m/s^2$.

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