wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

The rear side of a truck is open and a box of mass 20 kg is placed on the truck 4 m away from the open end. The coefficient of friction between the box and the surface is 0.15. The truck starts from rest with an acceleration of 2 m s2 on a straight road. The box will fall off the truck when it is at a distance from the starting point equal to (take g=10 ms2)

A
4 m
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
8 m
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
16 m
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
32 m
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C 16 m
Maximum acceleration of box =μg
=0.15×10 ms2=1.5 m s2
Acceleration of truck, aT=2 m s2
Therefore, relative acceleration of the box,
ar=0.5 m s2 (backwards)
It will fall off the truck in a time,
t=2Sar=2×40.5=4 s
Displacement of truck upto the instant is,
ST=12aTt2=12×2×(4)2=16 m

flag
Suggest Corrections
thumbs-up
46
similar_icon
Similar questions
View More
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Problem Solving Using Pseudo-Forces
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon