A block is kept inside an elevator which is ascending up at an angle 53⁰ with the horizontal with an acceleration of $5 m / s^{2} .$ Consider $g = 10 m / s^{2},$ the minimum value of coefficient of friction for which the block does not slip on the elevator floor is $μ .$ Find the value of $14 \times μ .$

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

answer is 3.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

Assuming the frictional force to act at its maximum,
$N - m (10 m / s^{2}) = m (5 m / s^{2}) \sin 53^{o}$
And $μ N = m (5 m / s^{2}) \cos 53^{o}$ we get $μ = \frac{3}{14}$

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET

How to Join IIT after 10th

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Get Expert Academic Guidance – Connect with a Counselor Today!