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
+1 vote
13.2k views
in Physics by (79.4k points)
closed by
A block of mass `m=5kg` is resting on a rough horizontal surface for which the coefficient of friction is `0.2` . When a force `F=40N` is applied, the acceleration of the block will be `(g=10m//s^(2))` .
image
A. `5.73m//sec^(2)`
B. `8.0m//sec^(2)`
C. `3.17m//sec^(2)`
D. `10.0m//sec^(2)`

1 Answer

0 votes
by (86.6k points)
selected by
 
Best answer
Correct Answer - A
Kinetic friction `=mu_(k)R=0.2(mg-Fsin30^(@))`
`=0.2(5xx10-40xx(1)/(2))= 0.2(50-20)=6N`
Acceleration of the block
`=(Fcos30^(@)-Ki n etic f riction)/(Mass)`
`=(40xxsqrt(3)/(2)-6)/(5)=5.73m//s^(2)` . image

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

...