Use app×
QUIZARD
QUIZARD
JEE MAIN 2026 Crash Course
NEET 2026 Crash Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
1.2k views
in Physics by (48.2k points)
closed by

An accelerated system with a vertical wall has co-efficient of friction µ between block and walls as shown in the figure. A block M of mass 1 kg just remains in equilibrium with the vertical wall, when the system has an acceleration of 20 m/s2. The co-efficient of friction has a value

(A) 0.10

(B) 0.25

(C) 0.50

(D) 1

1 Answer

+1 vote
by (49.3k points)
selected by
 
Best answer

Correct option is (C) 0.50

According to the problem the mass of the block is m = 1 kg

The acceleration of the cart is a = 20 m/s2

In this problem a force is working on the opposite direction of the acceleration

Let the force be f = mg

Therefore from newton's 2nd law,

N = ma

Let the coefficient friction of block be μ

Now the frictional force will be,

f = μN

mg = μ x ma

g = μ a

μ = g/a

If we assume g = 10 m/s2

therefore μ = 10/20

μ = 0.5

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.

...