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
134 views
in General by (85.8k points)
closed by
A box weight 1000 N is placed on the ground. The coefficient of friction between the box and the ground is 0.5. When the box is pulled by a 100 N horizontal force, the frictional force developed between the box and the ground at impending motion is
1. 50 N
2. 75 N
3. 100 N
4. 500 N

1 Answer

0 votes
by (96.5k points)
selected by
 
Best answer
Correct Answer - Option 3 : 100 N

Concept:

Frictional force resists the motion of the object upon a surface, however, this resisting force is limited, and once the applied force on the body is more than the resisting force the body will start moving.

The value of limiting friction is given by

Limiting friction F = μN

where μ is static friction coefficient between the body and the surface and is the N is the normal reaction of the body to the surface.

Calculation:

Given:

W = 1000 N, μ = 0.5, P = 100 N

Limiting friction force

F = μN

F = 0.5 × 1000 = 500 N

Since the applied force (100 N) is less than limiting friction so the body will be at rest.

 As long as the body is at rest, the friction force between the surface and the body will be equal to the force applied on the body

∴ Friction force = P = 100 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

...