Car is accelerating with acceleration a = 20m/ s². A box of mass m =10kg that is placed inside the car, is put in contact with the vertical wall of car as shown. The friction coefficient between the box and the wall is µ = 0.6

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

The net contact force between the vertical wall and the box is only of electromagnetic in nature.

The friction force acting on the box will be 100N

The acceleration of the box will be 20m/ s²

The contact force between the vertical wall and the box will be 100 $\sqrt{5}$ N

answer is A, B, C, D.

(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

w.r.t car
Question Image
So,
Since, mg is less than $f_{L} = 120 N$ , friction between the box and car will be 100N (static).
So, f=100 N and the box will not slide down the vertical wall. Contact force
$= \sqrt{N^{2} + f^{2}} = \sqrt{(200)^{2} + (100)^{2}} = 100 \sqrt{5} N$