JEE 2010 Study Plan for Physics 20 September

20 September Physics Friction – Ch.6 Session 9

Based on Text: Concepts of Physics, H C Verma, Part I

Do exercise problems 21 to 25

Before doing problems do a quick revision of the chapter and formulas.

Friction chapter formula revision sheet
http://iit-jee-physics.blogspot.com/2008/02/iit-jee-physics-formula-revision-6.html

Friction Past JEE Problem

1. A block of mass 1 kg lies on a horizontal surface in a truck. The coefficient of static friction between the block and the surface is 0.6. If the acceleration of the truck is 5 m/s², the frictional force acting on the block is _________ .
(JEE 1984)

Answer: 5 N

Reason: The horizontal force acting on the cube due to the acceleration of the truck is

F = ma = 1 kg * (5 m/s²) = 5 N

As the block is not moving, a frictional force of 5 N is acting on it.
We can verify that the static friction can go up to a maximum value of 1 kg *(10 m/s²)*0.6 = 6 N if we take g as approximately equal to 10 m/s².



From earlier chapter
Revise problems marked as difficult
If you have past JEE problems book go through problems related to the chapter.


Ask doubts if any in orkut community

IIT-JEE-Academy

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