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
11.3k views
in Physics by (64.2k points)

A block of mass 20 kg is acted upon by a force F = 30 N at an angle 53° with the horizontal in downward direction as shown. The coefficient of friction between the block and the horizontal surface is 0.2. The friction force acting on the block by the ground is (g = 10 m/s2)

(A)  40.0 N

(B)  30.0 N

(C)  18.0 N

(D)  44.8 N 

1 Answer

0 votes
by (61.3k points)
selected by
 
Best answer

Correct option (C) 18.0 N

Explanation:

Max. frictional force

 =  fmax = μ

μ(mg + F sin 53°)

= 0.2 (20 x 10 + 30 x 4/5)

= 44.8 N

As applied horizontal force is Fcos53° 

= 18N < fmax , friction force will also be 18 N.

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

...