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: JEE Main 2026 Registration Link | Top 30 Most Repeated Questions
JEE Main Prep: Study Plan | Preparation Tips | High Scoring Chapters and Topics
JEE Main QP & Mock: Previous 10 Year Questions | Chapter Wise PYQs | Mock test Series
JEE Main Most Scoring Concept: January 2025 Session | April 2025 Session
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 BrochureYour 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 BrochureYour Joint Entrance Exam Advanced brochure has been successfully mailed to your registered email id “”.
National Eligibility cum Entrance Test
Eligibility | Application | Result | Cutoff | College Predictor | Counselling | Answer Key | Accepting Colleges
Get Updates BrochureYour National Eligibility cum Entrance Test brochure has been successfully mailed to your registered email id “”.




