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
+8 votes
143k views
in Physics by (106k points)
edited by

Two masses m1 = 5kg and m2 = 10kg, connected by an inextensible string over a frictionless pulley, are moving as shown in the figure. The coefficient of friction of horizontal surface is 0.15. The minimum weight m that should be put on top of m2 to stop the motion is -

(1) 27.3 kg 

(2) 43.3 kg 

(3) 10.3 kg 

(4) 18.3 kg

2 Answers

+6 votes
by (32.5k points)
selected by
 
Best answer

50 – T = 5 × a
T – 0.15 (m + 10) g = (10 + m)a
a = 0 for rest
50 = 0.15 (m + 10) 10
5 =3/20(m + 10)
100/3= m + 10
m = 23.3 kg

+3 votes
by (20.4k points)

m1g = μ(m2 + m)
5 = 0.15 (10 + m)
500/15 = 10 + m
m = 350/15 = 23.3

So, correct answer will be option (1)

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

...