Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Laws of Motion
0 votes

A man is 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 man's shoes and belt is 0.2, upto what acceleration of belt can the man continue to be stationary relative to the belt? Mass of man$=65 kg; g=9.8 m/s^2$

a)$ 65\; N ,\; a_0=1.96 ms^{-2}$ b)$ 40\; N ,\; a_0=2.62 ms^{-2}$ c)$ 40\; N ,\; a_0=1.96 ms^{-2}$ d)$ 65\; N ,\; a_0=2.62 ms^{-2}$

Can you answer this question?

1 Answer

0 votes
As the man is stationary on the belt acceleration of man=acceleration of belt
Net force on the man $=ma$
$\qquad=65 \times 1=65 N$
Coefficient of friction $\mu=0.2$
Limiting friction $=\mu mg$
For the man to be stationary with respect to belt accelerating with $'a_0'$ of belt
$ma_0=\mu mg$
$a_0=\mu g$
$\qquad=0.2 \times 9.8 $
$\qquad=1.96 ms^{-2}$
Hence a is the correct answer.
answered Jul 8, 2013 by meena.p
edited May 27, 2014 by lmohan717

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App