logo

Ask Questions, Get Answers

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

A block of mass $m=2 kg$ is resting on a rough inclined plane of inclination $30 ^{\circ}$ as shown. the coefficient of friction between the block and the plane $\mu=0.5$ . What minimum force F should be applied perpendicular to the plane on the block so that the block does not slip on the plane $(g=10 m/s^2)$

a) zero

b) 6.24 N

c) 4.34 N

d) 2.68 N

 

Can you answer this question?
 
 

1 Answer

0 votes
Let F be maximum force required .
Normal force $N=F+mg \cos \theta$
Since the block does not slip
$mg \sin \theta=\mu N=f$
$\qquad=\mu(F+mg \cos 30^{\circ})$
or $F=\large\frac{mg \sin 30}{\mu}$$-mg \cos 30$
$\qquad=\large\frac{2(10)(1/2)}{.5}$$-2(10)\bigg(\large\frac{\sqrt 3}{2}\bigg)$
$\qquad=20-17.32$
$\qquad=2.68 N$
Hence d is the correct answer

 

answered Jul 11, 2013 by meena.p
edited Jan 26, 2014 by meena.p
 

Related questions

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