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 car of mass $2 \times 10 ^3 kg$ is to travel over a turn of radius $0.75 km$ and slope of angle $5^{\circ}$ If coefficient of friction between wheel and road is 0.5 maximum speed the car can travel with out slipping is $(\tan 5^{\circ}=0.087)$

\[(a)\;9.8 m/s \quad (b)\;67 m/s \quad (c)\;0.75 m/s \quad(d)\;1 m/s \]
Can you answer this question?

1 Answer

0 votes
$ R \cos \theta- f\sin \theta=mg$
$ R \sin \theta+ f\cos \theta=\large\frac{mv^2}{r}$
$f=\mu R$
$=>\large\frac{v^2}{rg}=\frac{\sin \theta+\mu \cos \theta}{\cos \theta-\mu \sin \theta}$
$=>\large\frac{\tan \theta+\mu}{1-\mu \sin \theta}$
$v=\sqrt {\large\frac{750 \times 10(0.087 +0.5)}{1-(0.5 \times 0.087)}}$
$\quad= 67 \;m/s$
Hence b is the correct answer.


answered Jul 5, 2013 by meena.p
edited Jan 26, 2014 by meena.p

Related questions

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