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  • The diagram below shows a motionless person relative to a horizontal treadmill accelerating at   <img alt="1 \,

The diagram below shows a motionless person relative to a horizontal treadmill accelerating at   1 \, {m}/{s^2} What is net strength in a man? If the coefficient of static friction between the person’s shoes and the belt is , at what acceleration of the belt can the person remain stationary relative to the belt?  (Human Mass = 65kg ).

Option: 1

2\ {m}/{s^2}


Option: 2

4\ {m}/{s^2}


Option: 3

5 \, m/s^2


Option: 4

10\, m/s^2


Answers (1)

We know that
The mass of the person = m\ =\ 65\ kg
The acceleration of the belt = a\ =1{\ m}/{s^2}
The coefficient of static friction = (\mu s) =0.2 
Therefore, 

Using Newton's second law of motion, the net force acting on the person the force (F) is
F\ =\ ma\ =\ 65N
The person can now stand still relative to the belt until the net force on the person is less than or equal to the force of friction between him and the belt (fs). 

Therefore,
\left(f_s\right)_{max}=f_s

Or,            

ma_{max}=\mu_sN\ =\ \mu_smg
Therefore,

a_{max}=\ \mu_sg
Or,

0.2\times10\ =\ 2\, {m}/{s^2}

Therefore, the maximum acceleration of the belt to a position where a person can stand is  2\ {m}/{s^2}

Posted by

Kshitij

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