5.36 The rear side of a truck is open and a box of 40 kg mass is placed 5 m away from the open end as shown in Fig. 5.22. The coefficient of friction between the box and the surface below it is 0.15. On a straight road, the truck starts from rest and accelerates with 2 m s-2. At what distance from the starting point does the box fall off the truck? (Ignore the size of the box).

7 Views|Posted 8 months ago
Asked by Shiksha User
1 Answer
V
8 months ago

Force on the box, F = MA = 40 * 2 N = 80 N

Frictional force, Ff = ? mg = 0.15 *40*10= 60 N

Net force = F – Ff = 80 – 60 = 20 N

From the equation F = ma, we get the backward acceleration produced in the box

a = 20/40 = 0.5 m/s2

From the equation s = ut + 12at2 , to travel s = 5 m by the box to fall off from the truck, we

...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 ncert solutions class 11th 2023

physics ncert solutions class 11th 2023

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