Hey there! We receieved your request
Stay Tuned as we are going to contact you within 1 Hour
One of our academic counsellors will contact you within 1 working day.
Click to Chat
1800-5470-145
+91 7353221155
Use Coupon: CART20 and get 20% off on all online Study Material
Complete Your Registration (Step 2 of 2 )
Sit and relax as our customer representative will contact you within 1 business day
OTP to be sent to Change
A block weighing 20 N rests on a horizontal surface. The coeff of static friction b/w the block and surface is 0.04 and the coeff of kinetic friction is 0.20. How long is the friction force exerted on the bock.
As the Q say ,block is resting on the horizontal suraface.So unless some external horizontal force is applied on it there is no
frictional force on it.To move the block min. hor. force required would be equal to maximum static frictional force=(coeff. of st. fr.)*(normal reaction N exerted by surface on the block)
where N=weight of the block=mg
Until applied force is greater tha above block will not move.At any time frictional force on the block = force applied(upto maximum static frictional force ).
After that as block gets in motion kinetic frictional force will oppose the motion.Here again max. kin. fr. force=(coeff. of kin. fr.)*N
As applied force is already greater than this, the max. kin. fr. force is the only opposing force to the same & it will be so as long as applied force>max. kin. fr.force.Whenever it became equal or less block will stop & to get it in motion again you have to apply force>max. st. fr. force as stated above.
Get your questions answered by the expert for free
You will get reply from our expert in sometime.
We will notify you when Our expert answers your question. To View your Question
Win Gift vouchers upto Rs 500/-
Register Yourself for a FREE Demo Class by Top IITians & Medical Experts Today !