Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
842 views
in Physics by (84.3k points)
closed by
A body is moving is down an inclined plane of slope `37^@` the coefficient of friction between the body and the plane varies as `mu=0.3x`, where x is the distance traveled down the plane by the body. The body will have maximum speed. `(sin 37^@ =(3)/(5))` .
A. at `x= 1.16 m`
B. at `x= 2m`
C. at bottommost point of the plane
D. at `x = 2.5 m`

1 Answer

0 votes
by (82.1k points)
selected by
 
Best answer
Correct Answer - D
`F("net") =mg sin theta -mu ng cos theta`
`=mg sin theta-0.3 xmg cos theta`…. (i)
At maximum speed `F_("net") =0`. Because after this net force Eq.(i), `F_("net") =0` at
`x=(tantheta)/(0.3) =(3/4)/(0.3) =2.5 m`

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...