An inclined plane making an angle of 300 with the horizontal is placed in a uniform horizontal electric field of 100 Vm . A particle of mass 1kg and charge 0.01C is allowed to slide down from rest from height of 1m. If the coefficient of friction is 0.2 , find the time taken by the particle to reach the bottom ( in seconds ) . Take g=9.8 m/s2
see full answer
Your Exam Success, Personally Taken Care Of
1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.
An Intiative by Sri Chaitanya
answer is 0001.34.
(Unlock A.I Detailed Solution for FREE)
Best Courses for You
JEE
NEET
Foundation JEE
Foundation NEET
CBSE
Detailed Solution
electrostatic force, F=Eq N=Eq sin30o+mg cos300 = 8.99N Acceleration down the plane a=mg sin300−μN−Eq cos300m ⇒a = (1)(9.8) (12) - (0.2)(8.99) - (100)(0.01)(32) 1 ⇒a=2.236 m/s2 Displacement along the plane S=1sin300=2m Using s=ut+12at2 ⇒2=122.236t2 ⇒42.236=t2 ⇒t=1.34 sec