The Total solution for NCERT class 6-12
Figure 5.18 shows a man standing stationary withrespect 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 betweenthe man’s shoes and the belt is 0.2, up to what acceleration of the belt canthe man continue to be stationary relative to the belt? (Mass of the man = 65kg.)
Figure5.18
Mass of the man, m = 65 kg
Acceleration of the belt, a =1 m/s2
Coefficient of static friction, μ =0.2
The net force F, acting on theman is given by Newton’s second law of motion as:
=65 × 1 = 65 N
The man will continue to be stationary withrespect to the conveyor belt until the net force on the man is less than orequal to the frictional force fs, exerted by the belt, i.e.,
∴a‘ = 0.2 × 10 = 2 m/s2
Therefore,the maximum acceleration of the belt up to which the man can stand stationaryis 2 m/s2.