CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

A 4Kg block A is placed on the top of a 8Kg block B which rests on a smooth table. A just slips on B when a force of 12 Newton is applied on A. The maximum horizontal force F on A required to make both A and B move together is

A
18N
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
24N
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
36N
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
48N
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A 18N
When they both move together
a=FTotalMass=F12
For A
F12=4×F12Newtons 2nd Law
FF3=12
F=18N
978160_1070940_ans_cee2499b5da1418caa4172666134d84d.png

flag
Suggest Corrections
thumbs-up
1
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Measurement of Significant Figures
CHEMISTRY
Watch in App
Join BYJU'S Learning Program
CrossIcon