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? | Learn NCERT solution | Education portal Class 11 Physics | Study online Unit-5 Laws Of Motion



Q.25:- 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.)



 

 

Answer:-

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 the man is given by Newton’s second law of motion as:
Fnet = ma = 65 × 1 = 65 N
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 = fs
ma’ = μmg
a‘ = 0.2 × 10 = 2 m/s2
Therefore, the maximum acceleration of the belt up to which the man can stand stationary is 2 m/s2.