A block B of mass 5kg is placed on a slab A of mass 20kg which lies on a frictionless surface as shown in the figure. The coefficient of static friction between the block and the slab is 0.4 and that of kinetic friction is 0.2. If a force F = 25N acts on B, the acceleration of the slab will be

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

zero

$1 m s^{- 2}$

$0.4 m s^{- 2}$

$0.5 m s^{- 2}$

answer is C.

(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

The maximum static friction is 20N

In this case both of the blocks will move together.

The combined acceleration will be $a = \frac{F}{m + M} = \frac{25}{5 + 20} = 1 {ms}^{- 2}$

Watch 3-min video & get full concept clarity

JEE

NEET

Foundation JEE

Foundation NEET

CBSE