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 ?
JEE Main 2025: Rank Predictor | College Predictor | Marks vs Rank vs Percentile
JEE Main 2025 Memory Based Question: Jan 22, 23, 24, 28 & 29 (Shift 1 & 2)
JEE Main 2025: High Scoring Topics | Sample Papers | Mock Tests | PYQs
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 | Answer Key | Result | Cutoff | College Predictor | Counselling | Application | Accepting Colleges
Get Updates BrochureYour National Eligibility cum Entrance Test brochure has been successfully mailed to your registered email id “”.