Two blocks (m = 0.5kg and M = 4.5kg) are arranged on a horizontal frictionless table as shown in figure. The coefficient of static friction between the two blocks is 3/7. Then the maximum horizontal force that can be applied on the block so that blocks move together is ---------N. (Round off to the Nearest Integer) [ Take g as 9.8 ms-2]

11 Views|Posted 5 months ago
Asked by Shiksha User
1 Answer
V
5 months ago

For the combined system of mass M and m, the acceleration under an applied force F is:
a = F / (M + m)

The static friction force (f_s) on the top block (m) provides its acceleration:
f_s = MA = m * [F / (M + m)] = mF / (M + m)

For the top block not to slip, the required static friction must be less than

...Read more

Thumbs Up IconUpvote Thumbs Down Icon

Taking an Exam? Selecting a College?

Get authentic answers from experts, students and alumni that you won't find anywhere else.

On Shiksha, get access to

66K
Colleges
|
1.2K
Exams
|
6.9L
Reviews
|
1.8M
Answers

Learn more about...

Physics Laws of Motion 2025

Physics Laws of Motion 2025

View Exam Details

Most viewed information

Summary

Share Your College Life Experience

Didn't find the answer you were looking for?

Search from Shiksha's 1 lakh+ Topics

or

Ask Current Students, Alumni & our Experts

Have a question related to your career & education?

or

See what others like you are asking & answering