If coefficient of friction is 0.5 , Applied force on a block is 49 N and mass of block is 10 kg then acceleration is ?
New: Direct link to apply for JEE Main 2025 registration for session 1
Also Check: Crack JEE Main 2025 - Join Our Free Crash Course Now!
JEE Main 2025: Sample Papers | Syllabus | Mock Tests | PYQs | Video Lectures
JEE Main 2025: Preparation Guide | High Scoring Topics | Study Plan 100 Days
We have = 0.5
f=N
Where N = mg
= 0.51010=50N
So, net force acting on the block in horizontal direction = 50N-49 N = 1 N
So, Acceleration of the block = Net force / mass of block = 1 N/10 kg = 0.1 meter per second square.
I hope that's clear.
Related Questions
Know More about
Joint Entrance Examination (Main)
Eligibility | Application | Preparation Tips | Question Paper | Admit Card | Answer Key | Result | Accepting Colleges
Get Updates Brochure
Your Joint Entrance Examination (Main) brochure has been successfully mailed to your registered email id “”.
Joint Entrance Exam Advanced
Eligibility | Application | Exam Pattern | Admit Card | Preparation Tips | Answer Key | Result | Accepting Colleges
Get Updates Brochure
Your Joint Entrance Exam Advanced brochure has been successfully mailed to your registered email id “”.
National Eligibility cum Entrance Test
Eligibility | Answer Key | Result | Cutoff | College Predictor | Counselling | Application | Accepting Colleges
Get Updates Brochure
Your National Eligibility cum Entrance Test brochure has been successfully mailed to your registered email id “”.
Upcoming Exams
Trending Articles around JEE Main
